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

[Alexnb]

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....
--
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.

[Mike Driscoll]

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

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

[Alexnb]

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


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.

[Alexnb]

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 "\\"?

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.

[Thomas Morton]

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

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" <alexnbryan@gma il.com>
Sent: Tuesday, June 10, 2008 7:05 PM
To: <python-list@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 "\\"?
>
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\Rhapsod y\Bryanbros\Wee zer\(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\Rhapso dy\Bryanbros\Ja son 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\Rhapsod y\Bryanbros\Jas on Mraz\I'm Yours (Single)\01 - I'm
>>Yours.wma")
>>>
>>WindowsErro r: [Error 2] The system cannot find the file specified:
>>"C:\\Document s and Settings\\Alex\ \My Documents\\My
>>Music\\Rhapso dy\\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
>--
>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-tp17759531p17761338.html

Sent from the Python - python-list mailing list archive at Nabble.com.
>
--
http://mail.python.org/mailman/listinfo/python-list

[Alexnb]

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" <alexnbryan@gma il.com>
Sent: Tuesday, June 10, 2008 7:05 PM
To: <python-list@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(rvar iable) 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\Rhapso dy\Bryanbros\We ezer\(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.startfil e("C:\Documen ts and Settings\Alex\M y Documents\My
>>>>Music\Rhaps ody\Bryanbros\J ason 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\Rhapso dy\Bryanbros\Ja son 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\\Rhaps ody\\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
>>--
>>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

--

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.

[Mike Driscoll]

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

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

[Alexnb]

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


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, 1:25 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
--

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.

[Thomas Morton]

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

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" <alexnbryan@gma il.com>
Sent: Tuesday, June 10, 2008 7:37 PM
To: <python-list@python.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" <alexnbryan@gma il.com>
>Sent: Tuesday, June 10, 2008 7:05 PM
>To: <python-list@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(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
>>>>
>>>>
>>>
>>--
>>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

>>
>--
>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-tp17759531p17761946.html

Sent from the Python - python-list mailing list archive at Nabble.com.
>
--
http://mail.python.org/mailman/listinfo/python-list

[Carsten Haese]

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

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

[Mike Driscoll]

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

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

[Mike Driscoll]

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

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

[Alexnb]

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


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, 1:57 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
--

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.

[Grant Edwards]

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

On 2008-06-11, Alexnb <alexnbryan@gma il.comwrote:

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.

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...

[Alexnb]

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


Okay, I don't understand how it is too vague, but here:

>>path = "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"
>>os.startfile( path)

Traceback (most recent call last):
File "<pyshell#3 9>", line 1, in <module>
os.startfile(pa th)
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"

Here's another way:

>>os.startfile( r"%s"%path)

Traceback (most recent call last):
File "<pyshell#4 0>", line 1, in <module>
os.startfile(r" %s"%path)
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"

Same output, however if I personally input it like so:

>>os.startfile( "C:\\Docume nts and Settings\\Alex\ \My Documents\\My
>>Music\\Rhapso dy\\Bryanbros\\ Jason Mraz\\I'm Yours (Single)\\01 - I'm
>>Yours.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 <alexnbryan@gma il.comwrote:
>

>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.

>
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.