Gerhard Häring wrote:
Alexnb wrote:
>Okay, so what I want my program to do it open a file, a music file in
specific, and for this we will say it is an .mp3. Well, I am using the
system() command from the os class. [...]
system("\"C:\D ocuments and Settings\Alex\M y Documents\My
Music\Rhapsody \Bryanbros\Weez er\(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:\Documen ts and Settings\Alex\M y Documents\My
Music\Rhapsod y\Bryanbros\Jas on Mraz\I'm Yours (Single)\01 - I'm
Yours.wma")
Traceback (most recent call last):
File "<pyshell#1 0>", line 1, in <module>
os.startfile("C :\Documents and Settings\Alex\M y 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:\\Docume nts and Settings\\Alex\ \My Documents\\My
Music\\Rhapsody \\Bryanbros\\Ja son 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....
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On Jun 10, 11:45*am, Alexnb <alexnbr...@gma il.comwrote:
Gerhard Häring wrote:
Alexnb wrote:
Okay, so what I want my program to do it open a file, a music file in
specific, and for this we will say it is an .mp3. Well, I am using the
system() command from the os class. [...]
system("\"C:\Do cuments and Settings\Alex\M y Documents\My
Music\Rhapsody\ Bryanbros\Weeze r\(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\M y Documents\My
Music\Rhapsody \Bryanbros\Jaso n Mraz\I'm Yours (Single)\01 - I'm
Yours.wma")
Traceback (most recent call last):
* File "<pyshell#1 0>", line 1, in <module>
* * os.startfile("C :\Documents and Settings\Alex\M y 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:\\Docume nts and Settings\\Alex\ \My Documents\\My
Music\\Rhapsody \\Bryanbros\\Ja son 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\f ile')
HTH
Mike
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 <alexnbr...@gma il.comwrote:
>Gerhard Häring wrote:
Alexnb wrote:
Okay, so what I want my program to do it open a file, a music file in
specific, and for this we will say it is an .mp3. Well, I am using the
system() command from the os class. [...]
>system("\"C:\D ocuments and Settings\Alex\M y Documents\My
Music\Rhapsody \Bryanbros\Weez er\(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:\Documen ts and Settings\Alex\M y Documents\My
Music\Rhapsod y\Bryanbros\Jas on Mraz\I'm Yours (Single)\01 - I'm
Yours.wma")
Traceback (most recent call last):
File "<pyshell#1 0>", line 1, in <module>
os.startfile("C :\Documents and Settings\Alex\M y Documents\My
Music\Rhapsody \Bryanbros\Jaso n Mraz\I'm Yours (Single)\01 - I'm
Yours.wma")
WindowsError : [Error 2] The system cannot find the file specified:
"C:\\Documen ts and Settings\\Alex\ \My Documents\\My
Music\\Rhapsod y\\Bryanbros\\J ason 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\f ile')
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.
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(rvari able) 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 <alexnbr...@gma il.comwrote:
>Gerhard Häring wrote:
Alexnb wrote:
Okay, so what I want my program to do it open a file, a music file in
specific, and for this we will say it is an .mp3. Well, I am using the
system() command from the os class. [...]
>system("\"C:\D ocuments and Settings\Alex\M y Documents\My
Music\Rhapsody \Bryanbros\Weez er\(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:\Documen ts and Settings\Alex\M y Documents\My
Music\Rhapsod y\Bryanbros\Jas on Mraz\I'm Yours (Single)\01 - I'm
Yours.wma")
Traceback (most recent call last):
File "<pyshell#1 0>", line 1, in <module>
os.startfile("C :\Documents and Settings\Alex\M y Documents\My
Music\Rhapsody \Bryanbros\Jaso n Mraz\I'm Yours (Single)\01 - I'm
Yours.wma")
WindowsError : [Error 2] The system cannot find the file specified:
"C:\\Documen ts and Settings\\Alex\ \My Documents\\My
Music\\Rhapsod y\\Bryanbros\\J ason 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\f ile')
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.
On Jun 10, 1:25*pm, "Thomas Morton" <morton.tho...@ googlemail.com>
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\Des ktop\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" <alexnbr...@gma il.com>
Sent: Tuesday, June 10, 2008 7:05 PM
To: <python-l...@python.org >
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(rvari able) 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 "\\"?
<snip>
Mike
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" <morton.tho...@ googlemail.com>
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\Des ktop\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" <alexnbr...@gma il.com>
Sent: Tuesday, June 10, 2008 7:05 PM
To: <python-l...@python.org >
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(rvari able) 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 "\\"?
<snip>
Mike
--
http://mail.python.org/mailman/listinfo/python-list
--
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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" <al********@gma il.com>
Sent: Tuesday, June 10, 2008 7:37 PM
To: <py*********@py thon.org>
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:\\Docume nts and Settings\\Alex\ \My Documents\\My
Music\\Rhapsody \\Bryanbros\\Ja son 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" <al********@gma il.com>
Sent: Tuesday, June 10, 2008 7:05 PM
To: <py*********@py thon.org>
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(rva riable) 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 <alexnbr...@gma il.comwrote:
Gerhard Häring wrote:
>
Alexnb wrote:
Okay, so what I want my program to do it open a file, a music file
in
specific, and for this we will say it is an .mp3. Well, I am using
the
system() command from the os class. [...]
>
system("\"C :\Documents and Settings\Alex\M y Documents\My
Music\Rhaps ody\Bryanbros\W eezer\(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.startfi le("C:\Document s and Settings\Alex\M y Documents\My
>Music\Rhap sody\Bryanbros\ Jason Mraz\I'm Yours (Single)\01 - I'm
>Yours.wma" )
>
Traceback (most recent call last):
File "<pyshell#1 0>", line 1, in <module>
os.startfile("C :\Documents and Settings\Alex\M y Documents\My
Music\Rhaps ody\Bryanbros\J ason Mraz\I'm Yours (Single)\01 - I'm
Yours.wma ")
>
WindowsErro r: [Error 2] The system cannot find the file specified:
"C:\\Docume nts and Settings\\Alex\ \My Documents\\My
Music\\Rhap sody\\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\m y\file')
HTH
Mike
--
http://mail.python.org/mailman/listinfo/python-list
--
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http://www.nabble.com/problems-with-...p17761338.html
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--
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--
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--
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http://www.nabble.com/problems-with-...p17761946.html
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--
http://mail.python.org/mailman/listinfo/python-list
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:\\Docume nts and Settings\\Alex\ \My Documents\\My
Music\\Rhapsody \\Bryanbros\\Ja son 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
On Jun 10, 2:09*pm, Carsten Haese <carsten.ha...@ gmail.comwrote:
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:\\Docume nts and Settings\\Alex\ \My Documents\\My
Music\\Rhapsody \\Bryanbros\\Ja son 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.sour ceforge.net
Sometimes I get too eager to help and don't do enough mental
processing before answering.
Mike
On Jun 10, 1:57*pm, Alexnb <alexnbr...@gma il.comwrote:
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?
<snip>
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 This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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