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# Re: problems with opening files due to file's path

 P: n/a Gerhard HÃ¤ring wrote: Alexnb wrote: >Okay, so what I want my program to do it open a file, a music file inspecific, and for this we will say it is an .mp3. Well, I am using thesystem() command from the os class. [...]system("\"C:\Documents and Settings\Alex\My Documents\MyMusic\Rhapsody\Bryanbros\Weezer\(2001)\04 - Island In The Sun.wma\"")[...] Try os.startfile() instead. It should work better. -- Gerhard -- http://mail.python.org/mailman/listinfo/python-list No, it didn't work, but it gave me some interesting feedback when I ran it in the shell. Heres what it told me: >>os.startfile("C:\Documents and Settings\Alex\My Documents\MyMusic\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'mYours.wma") Traceback (most recent call last): File "", line 1, in os.startfile("C:\Documents and Settings\Alex\My Documents\My Music\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'm Yours.wma") WindowsError: [Error 2] The system cannot find the file specified: "C:\\Documents and Settings\\Alex\\My Documents\\My Music\\Rhapsody\\Bryanbros\\Jason Mraz\\I'm Yours (Single)\x01 - I'm Yours.wma" See it made each backslash into two, and the one by the parenthesis and the 0 turned into an x.... -- View this message in context: http://www.nabble.com/problems-with-...p17759825.html Sent from the Python - python-list mailing list archive at Nabble.com. Jun 27 '08 #1
34 Replies

 P: n/a On Jun 10, 11:45*am, Alexnb os.startfile("C:\Documents and Settings\Alex\My Documents\MyMusic\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'mYours.wma") Traceback (most recent call last): * File "", line 1, in * * os.startfile("C:\Documents and Settings\Alex\My Documents\My Music\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'm Yours.wma") WindowsError: [Error 2] The system cannot find the file specified: "C:\\Documents and Settings\\Alex\\My Documents\\My Music\\Rhapsody\\Bryanbros\\Jason Mraz\\I'm Yours (Single)\x01 - I'm Yours.wma" See it made each backslash into two, and the one by the parenthesis and the 0 turned into an x.... -- View this message in context:http://www.nabble.com/problems-with-...o-file%27s-pat... Sent from the Python - python-list mailing list archive at Nabble.com. Yeah. You need to either double all the backslashes or make it a raw string by adding an "r" to the beginning, like so: os.startfile(r'C:\path\to\my\file') HTH Mike Jun 27 '08 #2

 P: n/a Hey thanks!, both the raw and the double backslashes worked. You are a gentleman and a scholar. Mike Driscoll wrote: On Jun 10, 11:45 am, Alexnb Gerhard HÃ¤ring wrote: Alexnb wrote:Okay, so what I want my program to do it open a file, a music file inspecific, and for this we will say it is an .mp3. Well, I am using thesystem() command from the os class. [...] >system("\"C:\Documents and Settings\Alex\My Documents\MyMusic\Rhapsody\Bryanbros\Weezer\(2001)\04 - Island In The Sun.wma\"")[...] Try os.startfile() instead. It should work better. -- Gerhard --http://mail.python.org/mailman/listinfo/python-list No, it didn't work, but it gave me some interesting feedback when I ranitin the shell. Heres what it told me: >>os.startfile("C:\Documents and Settings\Alex\My Documents\MyMusic\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'mYours.wma") Traceback (most recent call last): File "", line 1, in os.startfile("C:\Documents and Settings\Alex\My Documents\MyMusic\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'mYours.wma")WindowsError: [Error 2] The system cannot find the file specified:"C:\\Documents and Settings\\Alex\\My Documents\\MyMusic\\Rhapsody\\Bryanbros\\Jason Mraz\\I'm Yours (Single)\x01 - I'mYours.wma"See it made each backslash into two, and the one by the parenthesis andthe0 turned into an x....--View this message incontext:http://www.nabble.com/problems-with-...o-file%27s-pat...Sent from the Python - python-list mailing list archive at Nabble.com. Yeah. You need to either double all the backslashes or make it a raw string by adding an "r" to the beginning, like so: os.startfile(r'C:\path\to\my\file') HTH Mike -- http://mail.python.org/mailman/listinfo/python-list -- View this message in context: http://www.nabble.com/problems-with-...p17761126.html Sent from the Python - python-list mailing list archive at Nabble.com. Jun 27 '08 #3

 P: n/a Well, now i've hit another problem, this time being that the path will be a variable, and I can't figure out how to make startfile() make it raw with a variable, if I put startfile(r variable), it doesn't work and startfile(rvariable) obviously won't work, do you know how to make that work or better yet, how to take a regular string that is given and make every single "\" into a double "\\"? Mike Driscoll wrote: On Jun 10, 11:45 am, Alexnb Gerhard HÃ¤ring wrote: Alexnb wrote:Okay, so what I want my program to do it open a file, a music file inspecific, and for this we will say it is an .mp3. Well, I am using thesystem() command from the os class. [...] >system("\"C:\Documents and Settings\Alex\My Documents\MyMusic\Rhapsody\Bryanbros\Weezer\(2001)\04 - Island In The Sun.wma\"")[...] Try os.startfile() instead. It should work better. -- Gerhard --http://mail.python.org/mailman/listinfo/python-list No, it didn't work, but it gave me some interesting feedback when I ranitin the shell. Heres what it told me: >>os.startfile("C:\Documents and Settings\Alex\My Documents\MyMusic\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'mYours.wma") Traceback (most recent call last): File "", line 1, in os.startfile("C:\Documents and Settings\Alex\My Documents\MyMusic\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'mYours.wma")WindowsError: [Error 2] The system cannot find the file specified:"C:\\Documents and Settings\\Alex\\My Documents\\MyMusic\\Rhapsody\\Bryanbros\\Jason Mraz\\I'm Yours (Single)\x01 - I'mYours.wma"See it made each backslash into two, and the one by the parenthesis andthe0 turned into an x....--View this message incontext:http://www.nabble.com/problems-with-...o-file%27s-pat...Sent from the Python - python-list mailing list archive at Nabble.com. Yeah. You need to either double all the backslashes or make it a raw string by adding an "r" to the beginning, like so: os.startfile(r'C:\path\to\my\file') HTH Mike -- http://mail.python.org/mailman/listinfo/python-list -- View this message in context: http://www.nabble.com/problems-with-...p17761338.html Sent from the Python - python-list mailing list archive at Nabble.com. Jun 27 '08 #4

 P: n/a maybe try string substitution... not sure if that's really the BEST way to do it but it should work startfile(r"%s"%variable) -------------------------------------------------- From: "Alexnb" Sent: Tuesday, June 10, 2008 7:05 PM To: Subject: Re: problems with opening files due to file's path > Well, now i've hit another problem, this time being that the path will be a variable, and I can't figure out how to make startfile() make it raw with a variable, if I put startfile(r variable), it doesn't work and startfile(rvariable) obviously won't work, do you know how to make that work or better yet, how to take a regular string that is given and make every single "\" into a double "\\"? Mike Driscoll wrote: >>On Jun 10, 11:45 am, Alexnb >Gerhard HÃ¤ring wrote:Alexnb wrote:Okay, so what I want my program to do it open a file, a music file inspecific, and for this we will say it is an .mp3. Well, I am usingthesystem() command from the os class. [...]system("\"C:\Documents and Settings\Alex\My Documents\MyMusic\Rhapsody\Bryanbros\Weezer\(2001)\04 - Island In The Sun.wma\"")[...]Try os.startfile() instead. It should work better.-- Gerhard--http://mail.python.org/mailman/listinfo/python-listNo, it didn't work, but it gave me some interesting feedback when I ranitin the shell. Heres what it told me:os.startfile("C:\Documents and Settings\Alex\My Documents\MyMusic\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'mYours.wma")Traceback (most recent call last): File "", line 1, in os.startfile("C:\Documents and Settings\Alex\My Documents\MyMusic\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'mYours.wma")WindowsError: [Error 2] The system cannot find the file specified:"C:\\Documents and Settings\\Alex\\My Documents\\MyMusic\\Rhapsody\\Bryanbros\\Jason Mraz\\I'm Yours (Single)\x01 - I'mYours.wma"See it made each backslash into two, and the one by the parenthesis andthe0 turned into an x....--View this message incontext:http://www.nabble.com/problems-with-...o-file%27s-pat...Sent from the Python - python-list mailing list archive at Nabble.com. Yeah. You need to either double all the backslashes or make it a rawstring by adding an "r" to the beginning, like so:os.startfile(r'C:\path\to\my\file')HTHMike--http://mail.python.org/mailman/listinfo/python-list -- View this message in context: http://www.nabble.com/problems-with-...p17761338.html Sent from the Python - python-list mailing list archive at Nabble.com. -- http://mail.python.org/mailman/listinfo/python-list Jun 27 '08 #5

 P: n/a No this time it perhaps gave me the worst of all heres what I entered, and the output >>startfile(r"%s"%full) ***full is the path*** startfile(r"%s"%full) WindowsError: [Error 2] The system cannot find the file specified: '"C:\\Documents and Settings\\Alex\\My Documents\\My Music\\Rhapsody\\Bryanbros\\Jason Mraz\\I\'m Yours (Single)\x01 - I\'m Yours.wma"' Thomas Morton wrote: maybe try string substitution... not sure if that's really the BEST way to do it but it should work startfile(r"%s"%variable) -------------------------------------------------- From: "Alexnb" Sent: Tuesday, June 10, 2008 7:05 PM To: Subject: Re: problems with opening files due to file's path >>Well, now i've hit another problem, this time being that the path will be avariable, and I can't figure out how to make startfile() make it raw with avariable, if I put startfile(r variable), it doesn't work andstartfile(rvariable) obviously won't work, do you know how to make that workor better yet, how to take a regular string that is given and make everysingle "\" into a double "\\"?Mike Driscoll wrote: >>>On Jun 10, 11:45 am, Alexnb ", line 1, in os.startfile("C:\Documents and Settings\Alex\My Documents\MyMusic\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'mYours.wma")WindowsError: [Error 2] The system cannot find the file specified:"C:\\Documents and Settings\\Alex\\My Documents\\MyMusic\\Rhapsody\\Bryanbros\\Jason Mraz\\I'm Yours (Single)\x01 - I'mYours.wma"See it made each backslash into two, and the one by the parenthesis andthe0 turned into an x....--View this message incontext:http://www.nabble.com/problems-with-...o-file%27s-pat...Sent from the Python - python-list mailing list archive at Nabble.com.Yeah. You need to either double all the backslashes or make it a rawstring by adding an "r" to the beginning, like so:os.startfile(r'C:\path\to\my\file')HTHMike--http://mail.python.org/mailman/listinfo/python-list -- View this message in context: http://www.nabble.com/problems-with-...p17761338.htmlSent from the Python - python-list mailing list archive at Nabble.com.--http://mail.python.org/mailman/listinfo/python-list -- http://mail.python.org/mailman/listinfo/python-list -- View this message in context: http://www.nabble.com/problems-with-...p17761946.html Sent from the Python - python-list mailing list archive at Nabble.com. Jun 27 '08 #6

 P: n/a On Jun 10, 1:25*pm, "Thomas Morton" wrote: maybe try string substitution... not sure if that's really the BEST way to do it but it should work startfile(r"%s"%variable) I concur. That should work. A slightly more in depth example (assuming Windows): os.startfile(r'C:\Documents and Settings\%s\Desktop\myApp.exe' % username) or os.startfile(r'C:\Program Files\%s' % myApp) Hopefully this is what you are talking about. If you were referring to passing in arguments, than you'll want to use the subprocess module instead. > -------------------------------------------------- From: "Alexnb" Sent: Tuesday, June 10, 2008 7:05 PM To: Subject: Re: problems with opening files due to file's path Well, now i've hit another problem, this time being that the path will be a variable, and I can't figure out how to make startfile() make it raw with a variable, if I put startfile(r variable), it doesn't work and startfile(rvariable) obviously won't work, do you know how to make that work or better yet, how to take a regular string that is given and make every single "\" into a double "\\"? Mike Jun 27 '08 #7

 P: n/a That would work, but not for what I want. See the file could be anywhere on the user's system and so the entire path will be unique, and that didn't work with a unique path. What is the subprocess module you are talking about? Mike Driscoll wrote: On Jun 10, 1:25Â*pm, "Thomas Morton" wrote: >maybe try string substitution... not sure if that's really the BEST waytodo it but it should workstartfile(r"%s"%variable) I concur. That should work. A slightly more in depth example (assuming Windows): os.startfile(r'C:\Documents and Settings\%s\Desktop\myApp.exe' % username) or os.startfile(r'C:\Program Files\%s' % myApp) Hopefully this is what you are talking about. If you were referring to passing in arguments, than you'll want to use the subprocess module instead. >>--------------------------------------------------From: "Alexnb" Sent: Tuesday, June 10, 2008 7:05 PMTo: Subject: Re: problems with opening files due to file's path Well, now i've hit another problem, this time being that the path will be a variable, and I can't figure out how to make startfile() make it raw with a variable, if I put startfile(r variable), it doesn't work and startfile(rvariable) obviously won't work, do you know how to make that work or better yet, how to take a regular string that is given and make every single "\" into a double "\\"? Mike -- http://mail.python.org/mailman/listinfo/python-list -- View this message in context: http://www.nabble.com/problems-with-...p17762276.html Sent from the Python - python-list mailing list archive at Nabble.com. Jun 27 '08 #8

 P: n/a heh thanks Mike - glad im not going mad :P Just tested locally in IDLE (I know I know!) and it works for me like this: >>test = os.path.join(os.getcwd(),"NEWS.txt")test 'D:\\Python25\\NEWS.txt' >>os.startfile(r"%s"%test) And the file opens... Does the file definitely exist? Tom -------------------------------------------------- From: "Alexnb" Sent: Tuesday, June 10, 2008 7:37 PM To: Subject: Re: problems with opening files due to file's path > No this time it perhaps gave me the worst of all heres what I entered, and the output >>>startfile(r"%s"%full) ***full is the path*** startfile(r"%s"%full) WindowsError: [Error 2] The system cannot find the file specified: '"C:\\Documents and Settings\\Alex\\My Documents\\My Music\\Rhapsody\\Bryanbros\\Jason Mraz\\I\'m Yours (Single)\x01 - I\'m Yours.wma"' Thomas Morton wrote: >>maybe try string substitution... not sure if that's really the BEST waytodo it but it should workstartfile(r"%s"%variable)--------------------------------------------------From: "Alexnb" Sent: Tuesday, June 10, 2008 7:05 PMTo: Subject: Re: problems with opening files due to file's path >>>Well, now i've hit another problem, this time being that the path willbeavariable, and I can't figure out how to make startfile() make it rawwithavariable, if I put startfile(r variable), it doesn't work andstartfile(rvariable) obviously won't work, do you know how to make thatworkor better yet, how to take a regular string that is given and make everysingle "\" into a double "\\"?Mike Driscoll wrote:On Jun 10, 11:45 am, Alexnb Alexnb wrote:Okay, so what I want my program to do it open a file, a music fileinspecific, and for this we will say it is an .mp3. Well, I am usingthesystem() command from the os class. [...]>system("\"C:\Documents and Settings\Alex\My Documents\MyMusic\Rhapsody\Bryanbros\Weezer\(2001)\04 - Island In TheSun.wma\"")[...]>Try os.startfile() instead. It should work better.>-- Gerhard>--http://mail.python.org/mailman/listinfo/python-list>No, it didn't work, but it gave me some interesting feedback when Iranitin the shell. Heres what it told me:>>os.startfile("C:\Documents and Settings\Alex\My Documents\My>Music\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'm>Yours.wma")>Traceback (most recent call last): File "", line 1, in os.startfile("C:\Documents and Settings\Alex\My Documents\MyMusic\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'mYours.wma")>WindowsError: [Error 2] The system cannot find the file specified:"C:\\Documents and Settings\\Alex\\My Documents\\MyMusic\\Rhapsody\\Bryanbros\\Jason Mraz\\I'm Yours (Single)\x01 - I'mYours.wma">See it made each backslash into two, and the one by the parenthesisandthe0 turned into an x....--View this message incontext:http://www.nabble.com/problems-with-...o-file%27s-pat...Sent from the Python - python-list mailing list archive at Nabble.com.Yeah. You need to either double all the backslashes or make it a rawstring by adding an "r" to the beginning, like so:os.startfile(r'C:\path\to\my\file')HTHMike--http://mail.python.org/mailman/listinfo/python-list --View this message in context:http://www.nabble.com/problems-with-...p17761338.htmlSent from the Python - python-list mailing list archive at Nabble.com.--http://mail.python.org/mailman/listinfo/python-list --http://mail.python.org/mailman/listinfo/python-list -- View this message in context: http://www.nabble.com/problems-with-...p17761946.html Sent from the Python - python-list mailing list archive at Nabble.com. -- http://mail.python.org/mailman/listinfo/python-list Jun 27 '08 #9

 P: n/a Alexnb wrote: No this time it perhaps gave me the worst of all heres what I entered, and the output >>>startfile(r"%s"%full) ***full is the path*** startfile(r"%s"%full) WindowsError: [Error 2] The system cannot find the file specified: '"C:\\Documents and Settings\\Alex\\My Documents\\My Music\\Rhapsody\\Bryanbros\\Jason Mraz\\I\'m Yours (Single)\x01 - I\'m Yours.wma"' Contrary to what other posters have asserted, doing the above can't make a difference. Putting 'r' in front of a string literal tells Python not to give backslashes in the string literal any special treatment. Since there are no backslashes in "%s", the 'r' does nothing. Therefore, (r"%s"%full) is the same as ("%s"%full), which is the same as (full), assuming that full is the name of a string. The real answer lies in fixing the code where you're assigning the pathname to 'full', which you haven't posted. Please post the code where you're assigning the pathname, or better yet, post the complete code you're running. -- Carsten Haese http://informixdb.sourceforge.net Jun 27 '08 #10

 P: n/a On Jun 10, 2:09*pm, Carsten Haese >startfile(r"%s"%full) * ****full is the path*** startfile(r"%s"%full) WindowsError: [Error 2] The system cannot find the file specified: '"C:\\Documents and Settings\\Alex\\My Documents\\My Music\\Rhapsody\\Bryanbros\\Jason Mraz\\I\'m Yours (Single)\x01 - I\'m Yours.wma"' Contrary to what other posters have asserted, doing the above can't make a difference. Putting 'r' in front of a string literal tells Python not to give backslashes in the string literal any special treatment. Since there are no backslashes in "%s", the 'r' does nothing. I assumed the OP was trying to do the string substitution within a path. If the OP is instead doing as you think, then you are quite correct. > Therefore, (r"%s"%full) is the same as ("%s"%full), which is the same as (full), assuming that full is the name of a string. The real answer lies in fixing the code where you're assigning the pathname to 'full', which you haven't posted. Please post the code where you're assigning the pathname, or better yet, post the complete code you're running. -- Carsten Haesehttp://informixdb.sourceforge.net Sometimes I get too eager to help and don't do enough mental processing before answering. Mike Jun 27 '08 #11

 P: n/a On Jun 10, 1:57*pm, Alexnb As Carsten pointed out, we don't really know what you're doing. Or at least, I don't. Why do you even want to do string substitution? How does the user navigate to the files that the user wants to open? Are you using a GUI or a command line interface? Anyway, Google is your friend. Searching for "python subprocess" gives you this: http://docs.python.org/lib/module-subprocess.html Mike Jun 27 '08 #12

 P: n/a I am using GUI, Tkinter to be exact. But regardless of how the path gets there, it needs to opened correctly. The problem I am running into is that the program receives a path of a file, either .wma or .mp3 and is supposed to open it. I run into problems when there is either a ")" or a number next to the backslash "\" in the file path. I am looking for a way to make it work with a variable, I can make it work when I physically type it in, but not with a variable that holds the path. Mike Driscoll wrote: On Jun 10, 1:57Â*pm, Alexnb That would work, but not for what I want. See the file could be anywhereonthe user's system and so the entire path will be unique, and that didn'twork with a unique path. What is the subprocess module you are talkingabout? As Carsten pointed out, we don't really know what you're doing. Or at least, I don't. Why do you even want to do string substitution? How does the user navigate to the files that the user wants to open? Are you using a GUI or a command line interface? Anyway, Google is your friend. Searching for "python subprocess" gives you this: http://docs.python.org/lib/module-subprocess.html Mike -- http://mail.python.org/mailman/listinfo/python-list -- View this message in context: http://www.nabble.com/problems-with-...p17767518.html Sent from the Python - python-list mailing list archive at Nabble.com. Jun 27 '08 #13

 P: n/a On 2008-06-11, Alexnb

 P: n/a Okay, I don't understand how it is too vague, but here: >>path = "C:\Documents and Settings\Alex\My Documents\MyMusic\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'mYours.wma"os.startfile(path) Traceback (most recent call last): File "", line 1, in os.startfile(path) WindowsError: [Error 2] The system cannot find the file specified: "C:\\Documents and Settings\\Alex\\My Documents\\My Music\\Rhapsody\\Bryanbros\\Jason Mraz\\I'm Yours (Single)\x01 - I'm Yours.wma" Here's another way: >>os.startfile(r"%s"%path) Traceback (most recent call last): File "", line 1, in os.startfile(r"%s"%path) WindowsError: [Error 2] The system cannot find the file specified: "C:\\Documents and Settings\\Alex\\My Documents\\My Music\\Rhapsody\\Bryanbros\\Jason Mraz\\I'm Yours (Single)\x01 - I'm Yours.wma" Same output, however if I personally input it like so: >>os.startfile("C:\\Documents and Settings\\Alex\\My Documents\\MyMusic\\Rhapsody\\Bryanbros\\Jason Mraz\\I'm Yours (Single)\\01 - I'mYours.wma") It works out fine because I can make each backslash doubles so it doesn't mess stuff up. So if I could take the path varible and make ever "\" into a "\\" then it would also work. Did I clarify? Grant Edwards wrote: > On 2008-06-11, Alexnb I am using GUI, Tkinter to be exact. But regardless of how thepath gets there, it needs to opened correctly. The problem Iam running into is that the program receives a path of a file,either .wma or .mp3 and is supposed to open it. I run intoproblems when there is either a ")" or a number next to thebackslash "\" in the file path. I am looking for a way to makeit work with a variable, I can make it work when I physicallytype it in, but not with a variable that holds the path. You're going to have to show us code and example input and output. Your description of the problem is far too vague for anybody to help you. -- Grant Edwards grante Yow! With YOU, I can be at MYSELF... We don't NEED visi.com Dan Rather... -- http://mail.python.org/mailman/listinfo/python-list -- View this message in context: http://www.nabble.com/problems-with-...p17768511.html Sent from the Python - python-list mailing list archive at Nabble.com. Jun 27 '08 #15

 P: n/a Alexnb wrote: Okay, I don't understand how it is too vague, but here: [snip a bunch of irrelevant examples...] Did I clarify? No. Earlier you wrote: >On 2008-06-11, Alexnb >I am using GUI, Tkinter to be exact. But regardless of how thepath gets there, it needs to opened correctly. This implies that the file doesn't get opened correctly if the file name is entered/chosen in the GUI. Yet, the examples you posted don't contain any GUI code whatsoever. They merely demonstrate that you don't have a firm grasp on how backslashes in string literals are treated. So, this begs the question, do you actually have any GUI code that is failing, or are you just worried, given the problems you had with string literals, that the GUI code you have yet to write will fail in the same way? If this is the case, you should just write the GUI code and try it. It might just work. Backslashes entered into a GUI text box are not treated the same as backslashes in a Python string literal. If, on the other hand, you do have some GUI code for getting the file name from the user, and that code is failing, then please, show us THAT CODE and show us how it's failing. -- Carsten Haese http://informixdb.sourceforge.net Jun 27 '08 #16

 P: n/a On 2008-06-11, Alexnb >>path = "C:\Documents and Settings\Alex\My Documents\MyMusic\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'mYours.wma" Your string doesn't contain what you think it does. Do a "print path". Hint: the string "\01" consists of a single byte who's value is 001 base 8. >>>os.startfile(path) Traceback (most recent call last): File "", line 1, in os.startfile(path) WindowsError: [Error 2] The system cannot find the file specified: "C:\\Documents and Settings\\Alex\\My Documents\\My Music\\Rhapsody\\Bryanbros\\Jason Mraz\\I'm Yours (Single)\x01 - I'm Yours.wma" Notice that the string in the error message contains \x01? That's the byte that got changed. Here's another way: >>>os.startfile(r"%s"%path) Traceback (most recent call last): File "", line 1, in os.startfile(r"%s"%path) WindowsError: [Error 2] The system cannot find the file specified: "C:\\Documents and Settings\\Alex\\My Documents\\My Music\\Rhapsody\\Bryanbros\\Jason Mraz\\I'm Yours (Single)\x01 - I'm Yours.wma" Same output, however if I personally input it like so: >>>os.startfile("C:\\Documents and Settings\\Alex\\My Documents\\MyMusic\\Rhapsody\\Bryanbros\\Jason Mraz\\I'm Yours (Single)\\01 - I'mYours.wma") It works out fine because I can make each backslash doubles so it doesn't mess stuff up. Right. So if I could take the path varible and make ever "\" into a "\\" then it would also work. I don't understand the part about the path variable. The backslash character is used in both C and Python as an escape character so that you can encode special values in string literals. For example: '\r' is a carriage return '\n' is a linefeed, \0nnn is a single byte with the octal value nnn, and so on. Microsoft's choice of '\' as a path separator was a terribly bad one (one of many made by Microsoft over the years). Most Windows system calls will accept forward slashes, so the easiest thing to do is usually just type the strings with forward slashes. -- Grant Edwards grante Yow! NOW do I get to blow at out the CANLDES?? visi.com Jun 27 '08 #17

 P: n/a I don't think you understand it doesn't matter how the variable gets there, the same code is run regardless, I have no problem with the GUI, but you asked, and so I told you. the code os.startfile(.... is run if there is a GUI or it is a console app. Carsten Haese-2 wrote: > Alexnb wrote: >Okay, I don't understand how it is too vague, but here: [snip a bunch of irrelevant examples...]Did I clarify? No. Earlier you wrote: >>On 2008-06-11, Alexnb

 P: n/a Alexnb wrote: I don't think you understand it doesn't matter how the variable gets there But it *does* matter. Compare this: pyfilename = "C:\Somewhere\01 - Some Song.mp3" pyprint filename C:\Somewhere - Some Song.mp3 To this: pyfilename = raw_input("Enter the filename: ") Enter the filename: C:\Somewhere\01 - Some Song.mp3 pyprint filename C:\Somewhere\01 - Some Song.mp3 Note that the "\01" in the first case seems to have disappeared, whereas in the second case it's preserved. Now, if you want us to help you, please post your ACTUAL code with a description of the ACTUAL problem. -- Carsten Haese http://informixdb.sourceforge.net Jun 27 '08 #19

 P: n/a On Jun 11, 10:07*am, Alexnb >print 'First Line\nSecond Line' First Line Second Line The second behavior of the escape sequence is to make special character (generally the backspace itself), lose their special meaning: >>print 'path\\file.txt' path\file.txt In some cases, you might sometimes want a path like this: 'path \nick.txt' if you do this: >>print 'path\nick.txt' path ick.txt because the \n is considered as a newline. Instead, you should do this: >>print 'path\\nick.txt' path\nick.txt or >>print r'path\nick.txt' path\nick.txt the r'' string is raw string, most of the magics of a regular string '' is lost for an r'' string. It allows you to avoid the need to escape the special characters. Raw string is usually used for re (regular expressions) and paths in Windows both of which uses the backslash character regularly. you first case of: system("\"C:\Documents and Settings\Alex\My Documents\My Music\Rhapsody \Bryanbros\Weezer\(2001)\04 - Island In The Sun.wma\"") is interpreted by python as this: "C:\Documents and Settings\Alex\My Documents\My Music\Rhapsody \Bryanbros\Weezer\(2001) - Island In The Sun.wma" Notice that the '\0' is substituted into '', because \0 is the escape sequence for null character. (Personally I think python should raise errors if any escape character that isn't followed by a valid escape sequence is found (such as \D, \A, \M, \M, \R, \B, \W, \(), but perhaps they're trying not to be too mean for newbies.) that should be correctly written like this: system(r'"C:\Documents and Settings\Alex\My Documents\My Music\Rhapsody \Bryanbros\Weezer\(2001)\04 - Island In The Sun.wma"') or: system('"C:\\Documents and Settings\\Alex\\My Documents\\My Music\ \Rhapsody\\Bryanbros\\Weezer\\(2001)\\04 - Island In The Sun.wma"') Now, to the next question: How if I want the path to come from a variable: path = "C:\Documents and Settings\Alex\My Documents\My Music\Rhapsody \Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'm Yours.wma" os.startfile(path) I can see in a glance why it doesn't work, the string is escaped by python, it should be written like this. path = r"C:\Documents and Settings\Alex\My Documents\My Music\Rhapsody \Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'm Yours.wma" os.startfile(path) OR: path = "C:\\Documents and Settings\\Alex\\My Documents\\My Music\ \Rhapsody\\Bryanbros\\Jason Mraz\\I'm Yours (Single)\\01 - I'm Yours.wma" os.startfile(path) In most GUI toolkits (including Tkinter) and raw_input() function, when you input a string (using the textbox, a.k.a Entry widget) it would automatically be escaped for you, so when you input 'path\path \file.txt', the GUI toolkit would convert it into 'path\\path\ \file.txt'. Jun 27 '08 #20

 P: n/a Lie wrote: In most GUI toolkits (including Tkinter) and raw_input() function, when you input a string (using the textbox, a.k.a Entry widget) it would automatically be escaped for you, so when you input 'path\path \file.txt', the GUI toolkit would convert it into 'path\\path\ \file.txt'. That's incorrect. If you enter text into a text box or in raw_input(), *no* conversion of backslashes is happening. A backslash entered in raw_input is just a backslash. A backslash entered in a textbox is just a backslash. A backslash read from a file is just a backslash. A "conversion" happens when you print the repr() of a string that was obtained from raw_input or from a text box, because repr() tries to show the string literal that would result in the contents, and in a string literal, a backslash is not (always) a backslash, so repr() escapes the backslashes: pytext = raw_input("Enter some text: ") Enter some text: This is a backslash: \ pyprint text This is a backslash: \ pyprint repr(text) 'This is a backslash: \\' As you can see, I entered a single backslash, and the string ends up containing a single backslash. Only when I ask Python for the repr() of the string does the backslash get doubled up. -- Carsten Haese http://informixdb.sourceforge.net Jun 27 '08 #21

 P: n/a On Jun 11, 9:14*pm, Carsten Haese

 P: n/a Okay, so as a response to all of you, I will be using the Entry() widget in Tkinter to get this path. and the repr() function just makes all my backslashes 4 instead of just 1, and it still screwes it up with the numbers and parenthesis is has been since the first post. Oh and I know all about escape characters, (\n,\b,\a,etc.) I can program C, not a lot, but enough to know that I like python better. Anyway, so far I tried all of your stuff, and it didn't work. infact, it puts backslashes in front of the "'" in some of the words, such as "I'm" goes to "I\'m." So I posted the code I will be using if you want to see the Tkinter code I can post it, but I don't see how it will help. Lie Ryan wrote: On Jun 11, 9:14Â*pm, Carsten Haese Lie wrote: In most GUI toolkits (including Tkinter) and raw_input() function, when you input a string (using the textbox, a.k.a Entry widget) it would automatically be escaped for you, so when you input 'path\path \file.txt', the GUI toolkit would convert it into 'path\\path\ \file.txt'. That's incorrect. If you enter text into a text box or in raw_input(),*no* conversion of backslashes is happening. A backslash entered inraw_input is just a backslash. A backslash entered in a textbox is justa backslash. A backslash read from a file is just a backslash. I know, but I thought it'd be easier for him to understand it like that and discover the real 'how-it-works' later. My guilt is that I forget to put a "this is not how it actually works behind the scene, but you can think of it like this at least for now since this model is easier to grasp (although misleading)". >A "conversion" happens when you print the repr() of a string that wasobtained from raw_input or from a text box, because repr() tries to showthe string literal that would result in the contents, and in a stringliteral, a backslash is not (always) a backslash, so repr() escapes thebackslashes:pytext = raw_input("Enter some text: ")Enter some text: This is a backslash: \pyprint textThis is a backslash: \pyprint repr(text)'This is a backslash: \\'As you can see, I entered a single backslash, and the string ends upcontaining a single backslash. Only when I ask Python for the repr() ofthe string does the backslash get doubled up.--Carsten Haesehttp://informixdb.sourceforge.net -- http://mail.python.org/mailman/listinfo/python-list -- View this message in context: http://www.nabble.com/problems-with-...p17782866.html Sent from the Python - python-list mailing list archive at Nabble.com. Jun 27 '08 #23

 P: n/a On 2008-06-11, Alexnb

 P: n/a I don't get why yall are being so rude about this. My problem is this; the path, as a variable conflicts with other characters in the path, creating escape characters I don't want, so I need a way to send the string to the os.startfile() in raw, or, with all the backslashes doubled. Thats it, I'll write some code of what it should work like, because I probably should have done that; but you don't have to act like I am retarded... that solves nothing. Grant Edwards wrote: > On 2008-06-11, Alexnb Okay, so as a response to all of you, I will be using the Entry() widgetinTkinter to get this path. OK. >and the repr() function just makes all my backslashes 4instead of just 1, and it still screwes it up with the numbersand parenthesis is has been since the first post. I've absolutely no clue why you would be using the repr() function. >Oh and I know all about escape characters, (\n,\b,\a,etc.) Apparently not. >I can program C, not a lot, but enough to know that I likepython better. Anyway, so far I tried all of your stuff, andit didn't work. To what does "it" refer? >infact, it puts backslashes in front of the"'" in some of the words, such as "I'm" goes to "I\'m." Again, "it" doesn't seem to have a concrete referant. >So I posted the code I will be using if you want to see theTkinter code I can post it, but I don't see how it will help. If you know what would help and what wouldn't, then you must know enough to fix your problems. So please do so and quit bothering the newgroup. -- Grant Edwards grante Yow! I want another at RE-WRITE on my CEASAR visi.com SALAD!! -- http://mail.python.org/mailman/listinfo/python-list -- View this message in context: http://www.nabble.com/problems-with-...p17786386.html Sent from the Python - python-list mailing list archive at Nabble.com. Jun 27 '08 #28

 P: n/a Okay, so I wrote some code of basically what I will be doing, only with exactly what I need for this part of the program but here you go: [code] from Tkinter import* import os class myApp: def __init__(self, parent): self.parent = parent self.baseContainer = Frame(self.parent) self.baseContainer.pack() self.e = Entry(self.baseContainer) self.e.bind("", self.entryEnter) self.e.pack() self.Button1 = Button(self.baseContainer, command = self.buttonClick) self.Button1.configure(text="Submit") self.Button1.pack() def buttonClick(self): print "Button1 was clicked" path = self.e.get() path = "\"" + path + "\"" print path #os.startfile(path) def entryEnter(self, event): print "Enter was hit in the entry box" self.buttonClick() root = Tk() myapp = myApp(root) root.mainloop() [code] Alexnb wrote: > I don't get why yall are being so rude about this. My problem is this; the path, as a variable conflicts with other characters in the path, creating escape characters I don't want, so I need a way to send the string to the os.startfile() in raw, or, with all the backslashes doubled. Thats it, I'll write some code of what it should work like, because I probably should have done that; but you don't have to act like I am retarded... that solves nothing. Grant Edwards wrote: >>On 2008-06-11, Alexnb >Okay, so as a response to all of you, I will be using the Entry() widgetinTkinter to get this path. OK. >>and the repr() function just makes all my backslashes 4instead of just 1, and it still screwes it up with the numbersand parenthesis is has been since the first post. I've absolutely no clue why you would be using the repr()function. >>Oh and I know all about escape characters, (\n,\b,\a,etc.) Apparently not. >>I can program C, not a lot, but enough to know that I likepython better. Anyway, so far I tried all of your stuff, andit didn't work. To what does "it" refer? >>infact, it puts backslashes in front of the"'" in some of the words, such as "I'm" goes to "I\'m." Again, "it" doesn't seem to have a concrete referant. >>So I posted the code I will be using if you want to see theTkinter code I can post it, but I don't see how it will help. If you know what would help and what wouldn't, then you mustknow enough to fix your problems. So please do so and quitbothering the newgroup.--Grant Edwards grante Yow! I want another at RE-WRITE on my CEASAR visi.com SALAD!!--http://mail.python.org/mailman/listinfo/python-list -- View this message in context: http://www.nabble.com/problems-with-opening-files-due-to-file%27s-path-tp17759531p17786712.html Sent from the Python - python-list mailing list archive at Nabble.com. Jun 27 '08 #29

 P: n/a On Wed, Jun 11, 2008 at 4:16 PM, Alexnb I posted the underlying code, but I haven't made the GUI code because if I can't get the underlying code right it doesn't matter, well in my eyes it doesn't but I am probably wrong. But it will look somehting like this: What you're missing is that all of the problems you're having with escape characters *only apply to string literals inside your python source code.* If you're getting the string from the console, you don't need to escape or change anything. If you're getting the string from a Tk text box, you don't need to escape or change anything. If you're getting the string from the filesystem, you don't need to escape or change anything. The only place you have to be careful is when you type out a literal string inside your .py source code. When you do that, you need to properly escape backslashes, either by putting them in raw strings, or by doubling up your backslashes. If you haven't already, reread the section of the python tutorial about string literals: http://docs.python.org/tut/node5.html. -- Jerry Jun 27 '08 #30

 P: n/a Alexnb wrote: I don't get why yall are being so rude about this. We're frustrated with your apparent inability to understand anything we're saying. My problem is this; the path, as a variable conflicts with other characters in the path, creating escape characters I don't want, so I need a way to send the string to the os.startfile() in raw, or, with all the backslashes doubled. No, no, no, no, NO! That is not your problem! You are drawing unfounded conclusions from the fact that you had to double up backslashes when the filename came from a string literal. The situation is entirely different when the filename comes from user input. No doubling up of backslashes is necessary when the filename comes from user input. TRUST ME! For simplicity, start with this code to convince yourself: import os filename = raw_input("Please enter a filename: ") os.startfile(filename) Once you get that to work, replace raw_input with a function that gets the filename from your GUI. Hope this helps, -- Carsten Haese http://informixdb.sourceforge.net Jun 27 '08 #31

 P: n/a Well, I don't understand why I don't need to change anything because say I run that code, which goes straight from the entry box to the startfile() function. It doesn't work with some of the paths, that is the whole problem. the problem is when I enter a path with those certain characters next to each other. Everyone seems like there is some huge thing that I am missing... and I don't know what it is, because when I run that code, with that path I was talking about, I get an error, and it shows me that it is because of the \0. Jerry Hill wrote: > On Wed, Jun 11, 2008 at 4:16 PM, Alexnb >I posted the underlying code, but I haven't made the GUI code because ifIcan't get the underlying code right it doesn't matter, well in my eyes itdoesn't but I am probably wrong. But it will look somehting like this: What you're missing is that all of the problems you're having with escape characters *only apply to string literals inside your python source code.* If you're getting the string from the console, you don't need to escape or change anything. If you're getting the string from a Tk text box, you don't need to escape or change anything. If you're getting the string from the filesystem, you don't need to escape or change anything. The only place you have to be careful is when you type out a literal string inside your .py source code. When you do that, you need to properly escape backslashes, either by putting them in raw strings, or by doubling up your backslashes. If you haven't already, reread the section of the python tutorial about string literals: http://docs.python.org/tut/node5.html. -- Jerry -- http://mail.python.org/mailman/listinfo/python-list -- View this message in context: http://www.nabble.com/problems-with-...p17787168.html Sent from the Python - python-list mailing list archive at Nabble.com. Jun 27 '08 #32

 P: n/a Alexnb wrote: Okay, so I wrote some code of basically what I will be doing, only with exactly what I need for this part of the program but here you go: [...] Finally... path = "\"" + path + "\"" That line of code is unnecessary. Delete it. -- Carsten Haese http://informixdb.sourceforge.net Jun 27 '08 #33

 P: n/a Haha, okay well sorry that I was being so stupid, but I get it now and I apoligize for causing you all the frustration. But I did get it to work finally. Carsten Haese-2 wrote: > Alexnb wrote: >I don't get why yall are being so rude about this. We're frustrated with your apparent inability to understand anything we're saying. >My problem is this; thepath, as a variable conflicts with other characters in the path, creatingescape characters I don't want, so I need a way to send the string to theos.startfile() in raw, or, with all the backslashes doubled. No, no, no, no, NO! That is not your problem! You are drawing unfounded conclusions from the fact that you had to double up backslashes when the filename came from a string literal. The situation is entirely different when the filename comes from user input. No doubling up of backslashes is necessary when the filename comes from user input. TRUST ME! For simplicity, start with this code to convince yourself: import os filename = raw_input("Please enter a filename: ") os.startfile(filename) Once you get that to work, replace raw_input with a function that gets the filename from your GUI. Hope this helps, -- Carsten Haese http://informixdb.sourceforge.net -- http://mail.python.org/mailman/listinfo/python-list -- View this message in context: http://www.nabble.com/problems-with-...p17787228.html Sent from the Python - python-list mailing list archive at Nabble.com. Jun 27 '08 #34

 P: n/a Alexnb wrote: Haha, okay well sorry that I was being so stupid, but I get it now and I apoligize for causing you all the frustration. But I did get it to work finally. Jun 27 '08 #35

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