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Variable passing to external program - How??

Hi,

Ok first please bear with me as I am a total Python n00b..

Can anyone explain why this does not like me using % FileLoc in the
os.system call???

#!/usr/bin/python
import sys
import os
# Check that the folder is accessible and writeable
FileLoc = os.path.exists( '/home/rigga')
if (FileLoc):
print "File location exists:"
AccFlag = os.access('% FileLoc',os.R_O K | os.X_OK | os.W_OK)
if (AccFlag):
print "You have FULL access to the location"
else:
print "**Error - You do not have access to the location**"
else:
print "No files found exiting..."

sys.exit()

I have full access to the folder I am checking however it always returns no
files found (FileLoc = 0) however if I specify the folder I want to test in
the os.system call it works fine...

Any help appreciated

Cheerz

Rigga

Jul 18 '05 #1
10 2532
Rigga wrote:
Hi,

Ok first please bear with me as I am a total Python n00b..
OK, but: the post's subject bears no relation to your code; and
neither does your text -- you keep talking about an os.system
call that just isn't there. So, "noob" or not, I'm nonplussed.
Can anyone explain why this does not like me using % FileLoc in the
os.system call???
There is no os.system call in the following code.
#!/usr/bin/python
import sys
import os
# Check that the folder is accessible and writeable
FileLoc = os.path.exists( '/home/rigga')
if (FileLoc):
print "File location exists:"
AccFlag = os.access('% FileLoc',os.R_O K | os.X_OK | os.W_OK)
You're checking for a file called '% FileLoc', which does not exist. The
variable FileLoc at this point is worth True, so you can't possibly want
to "pass it to an external program" as per subject, either.
I have full access to the folder I am checking however it always returns
no files found (FileLoc = 0) however if I specify the folder I want to
test in the os.system call it works fine...


There is no os.system call anywhere in the above.
Alex

Jul 18 '05 #2
> FileLoc = os.path.exists( '/home/rigga')

This yields true or false, depending on the existence of "/home/rigga"
I'm guessing now, but what you want is this:

FileLoc = "/home/rigga"
if os.path.exists( FileLoc):
....
if (FileLoc):
print "File location exists:"
AccFlag = os.access('% FileLoc',os.R_O K | os.X_OK | os.W_OK)
Guessing again - you want to check your perms on that FileLoc of yours - so
why don't you just pass it into the function?
AccFlag = os.access(FileL oc,os.R_OK | os.X_OK | os.W_OK)

The % operator works similar to printf in C/PHP/Whatever. If you absolutely
want it here, this would work:
print "%s" % FileLoc

"/home/rigga"
--
Diez
Jul 18 '05 #3
> #!/usr/bin/python
import sys
import os
# Check that the folder is accessible and writeable
FileLoc = os.path.exists( '/home/rigga')
if (FileLoc):
print "File location exists:"
AccFlag = os.access('% FileLoc',os.R_O K | os.X_OK | os.W_OK)
if (AccFlag):
print "You have FULL access to the location"
else:
print "**Error - You do not have access to the location**"
else:
print "No files found exiting..."

sys.exit()


import os
FileLoc = '/home/rigga'

if os.path.exists( FileLoc):
print "File location exists"
AccFlag = os.access(FileL oc, os.R_OK | os.X_OK | os.W_OK)
if AccFlag:
print "You have FULL access to the location"
else:
print "**Error - You do not have access to the location**"
else:
print "No files found exiting..."

Jul 18 '05 #4
Alex Martelli wrote:
Rigga wrote:
Hi,

Ok first please bear with me as I am a total Python n00b..


OK, but: the post's subject bears no relation to your code; and
neither does your text -- you keep talking about an os.system
call that just isn't there. So, "noob" or not, I'm nonplussed.
Can anyone explain why this does not like me using % FileLoc in the
os.system call???


There is no os.system call in the following code.
#!/usr/bin/python
import sys
import os
# Check that the folder is accessible and writeable
FileLoc = os.path.exists( '/home/rigga')
if (FileLoc):
print "File location exists:"
AccFlag = os.access('% FileLoc',os.R_O K | os.X_OK | os.W_OK)


You're checking for a file called '% FileLoc', which does not exist. The
variable FileLoc at this point is worth True, so you can't possibly want
to "pass it to an external program" as per subject, either.
I have full access to the folder I am checking however it always returns
no files found (FileLoc = 0) however if I specify the folder I want to
test in the os.system call it works fine...


There is no os.system call anywhere in the above.
Alex

Alex, point taken I rushed when I was doing this and made many mistakes,
thatll learn me not to rush a post!.

To all the others that replied thanks for your input it has helped me a lot.

What I was going to add to the original post but forgot was how you pass
variables to external programs - hence the title. I have been reading
though the Python Bible however all this variable passing doesnt appear to
work as I would expect i.e.

FilePath = os.path('/home/rigga')
AccFlag = os.access('% FilePath',os.R_ OK)

I would expect % FilePath to contain /home/rigga and for os.access to parse
% FilePath in to that and return the results.... however it doesnt, it only
works if I specify the directory in os.access...

Or am I just being stupid?

Cheerz

Rigga
Jul 18 '05 #5
On Sat, 11 Oct 2003 15:50:29 +0000, Rigga <Ri***@noemail. com> wrote:
FilePath = os.path('/home/rigga')
AccFlag = os.access('% FilePath',os.R_ OK)

I would expect % FilePath to contain /home/rigga


Why would you expect that?
--
Christopher
Jul 18 '05 #6
Christopher Koppler wrote:
On Sat, 11 Oct 2003 15:50:29 +0000, Rigga <Ri***@noemail. com> wrote:
FilePath = os.path('/home/rigga')
AccFlag = os.access('% FilePath',os.R_ OK)

I would expect % FilePath to contain /home/rigga


Why would you expect that?
--
Christopher

because Ive assigned it using the FilePath = os.path('/home/rigga') -
surely therefore FilePath contains the value /home/rigga???????
Jul 18 '05 #7
On Sat, 11 Oct 2003 16:08:46 +0000, Rigga <Ri***@noemail. com> wrote:
Christopher Koppler wrote:
On Sat, 11 Oct 2003 15:50:29 +0000, Rigga <Ri***@noemail. com> wrote:
FilePath = os.path('/home/rigga')
AccFlag = os.access('% FilePath',os.R_ OK)

I would expect % FilePath to contain /home/rigga


Why would you expect that?


because Ive assigned it using the FilePath = os.path('/home/rigga') -
surely therefore FilePath contains the value /home/rigga???????


Yes, but what do you think that '% FilePath' means? This is an
ordinary string, and does not magically expand to the variable's
contents, which I assume you wanted. You need to use just the variable
name for that:

AccFlag = os.access(FileP ath, os.R_OK)

If you wanted to use the % operator for strings, you could also write
that as

AccFlag = os.access('%s' % FilePath, os.R_OK)

which is completely unnecessary in this case, however.

--
Christopher
Jul 18 '05 #8
On Sat, 11 Oct 2003 15:22:36 GMT, Christopher Koppler
<kl******@chell o.at> wrote:
On Sat, 11 Oct 2003 16:08:46 +0000, Rigga <Ri***@noemail. com> wrote:
Christopher Koppler wrote:
On Sat, 11 Oct 2003 15:50:29 +0000, Rigga <Ri***@noemail. com> wrote:

FilePath = os.path('/home/rigga')
AccFlag = os.access('% FilePath',os.R_ OK)

I would expect % FilePath to contain /home/rigga

Why would you expect that?


because Ive assigned it using the FilePath = os.path('/home/rigga') -
surely therefore FilePath contains the value /home/rigga???????


Yes, but what do you think that '% FilePath' means? This is an
ordinary string, and does not magically expand to the variable's
contents, which I assume you wanted. You need to use just the variable
name for that:

AccFlag = os.access(FileP ath, os.R_OK)

If you wanted to use the % operator for strings, you could also write
that as

AccFlag = os.access('%s' % FilePath, os.R_OK)

which is completely unnecessary in this case, however.


And also, I completely overlooked:
FilePath = os.path('/home/rigga') will not work either, because
os.path is a _module_ (which is not callable), not a function to
create paths. Pathnames are just strings, so

FilePath = '/home/rigga'

is what you want.
--
Christopher
Jul 18 '05 #9
Following is yet another version for checking file access
that takes the file_path as an argument to the module,
and differentiates R,E,W or NO access ....

'''
Module ....... os_file_access. py
NewsGroup .... comp.lang.pytho n
Date ......... 2003-10-10
Posted_By .... rigga
Edited_By .... Stanley C. Kitching
'''

import os
import sys

NL = '\n'
SP = ' '
SP3 = SP * 3
SP7 = SP * 7

module_this = sys.argv[ 0 ]

print '%s %s ' % ( NL , module_this ) , NL

if len( sys.argv ) < 2 :

print SP7 , 'Usage : python os_file_access. py file_path' , NL

sys.exit( -1 )

file_path = sys.argv[ 1 ] # Input File Path as 1st Argument

if os.path.exists( file_path ) :

print SP7 , "File Path Exists ...." , file_path , NL

read_flag = os.access( file_path , os.R_OK )
exec_flag = os.access( file_path , os.X_OK )
write_flag = os.access( file_path , os.W_OK )

else :

print SP7 , "*** File NOT Found ***" , NL
print SP7 , SP3 , file_path , NL

sys.exit( -2 )

list_access = []

if read_flag : list_access.app end( 'Read' )
if exec_flag : list_access.app end( 'Execute' )
if write_flag : list_access.app end( 'Write' )

if ( read_flag | exec_flag | write_flag ) :

str_access = ' , '.join( list_access )

else :

str_access = 'NO'

print SP7 , 'You have [ %s ] Access to the File' % ( str_access ) , NL

--
Cousin Stanley
Human Being
Phoenix, Arizona

Jul 18 '05 #10

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