Re：由于文件路径而导致打开文件的问题 [英] Re: problems with opening files due to file's path
问题描述
Gerhard H？¤ring写道：
Alexnb写道：
>好的，所以我希望我的程序能够打开一个文件，特定的音乐文件，为此我们会说它是.mp3。好吧，我正在使用os类的
system（）命令。 [...]
系统（\C：\Documents and Settings \Alex \ My Documents \ Myy /> Music\Rhapsody\Bryanbros \\ Weezer \（2001）\ 04 - Island In The Sun.wma \""）
[...]
试试os.startfile（）代替。它应该会更好。
- Gerhard
-
http://mail.python.org/mailman/listinfo/python-list
不，它没有用，但是当我在shell中运行它时，它给了我一些有趣的反馈
。下面是它告诉我的：
>> os.startfile（" C： \ Documents and Settings\Alex\My Documents\My
Music\Rhapsody \ Bryanbros \ Jason Mraz \I''m Yours（Single）\ 01 - 我是
Yours.wma"）
Traceback（最近一次调用最后一次）：
文件" < pyshell＃10>"，第1行，在< module>
os.startfile（" C：\Documents and Settings \ Alex \ MyMy Documents \ Myy
Music\Rhapsody \ Bryanbros \ Jason Mraz;我是你的（单身）\ 01 - 我是Yours.wma"）
WindowsError：[错误2]系统找不到指定的文件：
" C：\\Documents and Settings \\Alex \\ Myy Documents \\我的
Music\\Rhapsody \\ Bryanbros \\\ Jason Mraz \\I''m Yours（Single）\ x01 - 我是
Yours.wma"
见它将每个反斜杠分成两个，括号中的那个和
0变成了一个x ....
-
在上下文中查看此消息： http://www.nabble.com/problems-with-...p17759825.html
从Nabble的Python-python-list邮件列表存档发送。 com。
6月10日11:45 * am，Alexnb< alexnbr ... @ gmail.comwrote：
Gerhard H？ring写道：
Alexnb写道：
好的，所以我希望我的程序能够打开一个文件，特定于
的音乐文件，为此我们会说它是.m P3。好吧，我正在使用os类中的
system（）命令。 [...]
system（" \" C：\Documents和设置\ Alex \我的文件\我/
音乐\狂想曲'\\ Bryanbros \ Weezer \（2001）\ 04 - Island In The Sun.wma \"" ;）
[...]
请尝试使用os.startfile（）。它应该更好。
- Gerhard
-
< a rel =nofollowhref =http://mail.python.org/mailman/listinfo/python-listtarget =_ blank> http://mail.python.org/mailman/listinfo/python-list
不，它没有用，但是当我在shell中运行它时，它给了我一些有趣的反馈
。下面是它告诉我的：
> os.startfile（" C：\Documents and Settings\Alex\我的文件\ MyMy
Music\Rhapsody \ Bryanbros \ Jason Mraz \我是你的（单身）\ 01 - 我是
Yours.wma"）
回溯（最近一次调用最后一次）：
*文件"< pyshell＃10>"，第1行，in < module>
* * os.startfile（" C：\Documents and Settings\Alex\My Documents\My
Music\Rhapsody \\ \\ Bryanbros \ Jason Mraz \我是你的（单身）\ 01 - 我是你的.wma"）
WindowsError：[错误2]系统不能找到指定的文件：
" C：\\Documents and Settings \\Alex \\ My Documents \\My
Music \\ \\\ Rhapsody \\Bryanbros \\ Jason Mraz \\I''Y You rs（Single）\ x01 - 我是
Yours.wma"
看到它将每个反斜杠分成两个，括号中的一个并且
0变成了x ....
-
在上下文中查看此消息： http://www.nabble.com/problems-with -... o-file％27s-pat ...
从Nabble.com的Python-python-list邮件列表存档发送。
是的。您需要将所有反斜杠加倍，或者通过添加r来使其成为原始的
字符串。一开始，就像这样：
os.startfile（r''C：\ path\to\my\file''）
>
HTH
Mike
嘿谢谢！，原始和双反斜杠工作。你是一个绅士和一个学者。
Mike Driscoll写道：
6月10日上午11：45，Alexnb< alexnbr。 .. @ gmail.comwrote：
> Gerhard H？¤ring写道：
Alexnb写道：
好的，所以我希望我的程序能够打开一个文件，一个特定的音乐文件，为此我们会说它是一个.mp3。好吧，我正在使用os类的
system（）命令。 [...]
> system（" \" C：\Documents and Settings \Alex\ My Documents \\ \\ Myy
Music\Rhapsody \ Bryanbros \ Weezer \（2001）\ 04 - Island In The Sun.wma \"
[...] $ ... $ b
请尝试使用os.startfile（）。它应该更好。
- Gerhard
-
< a rel =nofollowhref =http://mail.python.org/mailman/listinfo/python-listtarget =_ blank> http://mail.python.org/mailman/listinfo/python-list
不，它没有用，但是当我在shell中运行
它时，它给了我一些有趣的反馈。下面是它告诉我的内容：
>> os.startfile（" C：\Documents and Settings \ Alex \ MyMy Documents \ My
Music\Rhapsody \ Bryanbros \ Jason Mraz \我是你的（单身）\ 01 - 我是
Yours.wma"）
<回溯（最近一次调用最后一次）：
文件"< pyshell＃10>"，第1行，< module>
os.startfile（" C：\Documents和设置\ Alex \我的文件\我/
Music\Rhapsody \ Bryanbros \ Jason Mraz \我是你的（单身）\ 01 - 我是你的
你的.wma"）
WindowsError：[错误2]系统找不到指定的文件：
C：\\Documents and Settings \\Alex \\我的文件\\\\My
音乐\\Rhapsody \\Bryanbros \\ Jason Mraz \\I''m Yours（Single）\ x01 - 我是
Yours.wma"
看看每个反斜杠分为两个，一个由括号和
0变成一个x ....
-
在
上下文中查看此消息：< a rel =nofollowhref =http://www.nabble.com/problems-with-opening-files-due-to-file%27s-pattarget =_ blank> http：//www.nabble .com / problems-with -... o-file％27s-pat ...
从Nabble.com的Python-python-list邮件列表存档发送。
是的。您需要将所有反斜杠加倍，或者通过添加r来使其成为原始的
字符串。一开始，就像这样：
os.startfile（r''C：\ path\to\my\file''）
>
HTH
Mike
-
http://mail.python.org/mailman/listinfo/python-list
-
在上下文中查看此消息： http://www.nabble.com/problems-with-...p17761126.html
从Nabble.com的Python-python-list邮件列表存档发送。
嗯，现在我已经遇到另一个问题，这次是路径将是一个
变量，而我无法弄清楚如何使startfile（）使其成为原始的
变量，如果我把startfile（r变量），它不起作用和
startf ile（rvariable）显然不会工作，你知道如何使这项工作
或更好，如何获取一个给定的常规字符串并使每一个
单&\\成为双重的\\ -
Mike Driscoll写道：
On Jun 10点，11点45分，Alexnb< alexnbr ... @ gmail.comwrote：
> Gerhard H？¤ring写道：
Alexnb写道：
好的，所以我希望我的程序能够打开一个文件，一个特定的音乐文件，为此我们会说它是一个.mp3。好吧，我正在使用os类的
system（）命令。 [...]
> system（" \" C：\Documents and Settings \Alex\ My Documents \\ \\ Myy
Music\Rhapsody \ Bryanbros \ Weezer \（2001）\ 04 - Island In The Sun.wma \"
[...] $ ... $ b
请尝试使用os.startfile（）。它应该更好。
- Gerhard
-
< a rel =nofollowhref =http://mail.python.org/mailman/listinfo/python-listtarget =_ blank> http://mail.python.org/mailman/listinfo/python-list
不，它没有用，但是当我在shell中运行
它时，它给了我一些有趣的反馈。下面是它告诉我的内容：
>> os.startfile（" C：\Documents and Settings \ Alex \ MyMy Documents \ My
Music\Rhapsody \ Bryanbros \ Jason Mraz \我是你的（单身）\ 01 - 我是
Yours.wma"）
<回溯（最近一次调用最后一次）：
文件"< pyshell＃10>"，第1行，< module>
os.startfile（" C：\Documents和设置\ Alex \我的文件\我/
Music\Rhapsody \ Bryanbros \ Jason Mraz \我是你的（单身）\ 01 - 我是你的
你的.wma"）
WindowsError：[错误2]系统找不到指定的文件：
C：\\Documents and Settings \\Alex \\我的文件\\\\My
音乐\\Rhapsody \\Bryanbros \\ Jason Mraz \\I''m Yours（Single）\ x01 - 我是
Yours.wma"
看看每个反斜杠分为两个，一个由括号和
0变成一个x ....
-
在
上下文中查看此消息：< a rel =nofollowhref =http://www.nabble.com/problems-with-opening-files-due-to-file%27s-pattarget =_ blank> http：//www.nabble .com / problems-with -... o-file％27s-pat ...
从Nabble.com的Python-python-list邮件列表存档发送。
是的。您需要将所有反斜杠加倍，或者通过添加r来使其成为原始的
字符串。一开始，就像这样：
os.startfile（r''C：\ path\to\my\file''）
>
HTH
Mike
-
http://mail.python.org/mailman/listinfo/python-list
-
在上下文中查看此消息： http://www.nabble.com/problems-with-...p17761338.html
从Nabble.com的Python-python-list邮件列表存档发送。
Gerhard H?¤ring 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\My Documents\My
Music\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\My
Music\Rhapsody\Bryanbros\Jason Mraz\I''m Yours (Single)\01 - I''m
Yours.wma")
Traceback (most recent call last):
File "<pyshell#10>", line 1, in <module>
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.
On Jun 10, 11:45*am, Alexnb <alexnbr...@gmail.comwrote:Gerhard H?ring 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\My Documents\My
Music\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\My
Music\Rhapsody\Bryanbros\Jason Mraz\I''m Yours (Single)\01 - I''m
Yours.wma")
Traceback (most recent call last):
* File "<pyshell#10>", line 1, in <module>
* * 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
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...@gmail.comwrote:>Gerhard H?¤ring 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\My Documents\My
Music\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\My
Music\Rhapsody\Bryanbros\Jason Mraz\I''m Yours (Single)\01 - I''m
Yours.wma")
Traceback (most recent call last):
File "<pyshell#10>", line 1, in <module>
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
--
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(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 <alexnbr...@gmail.comwrote:>Gerhard H?¤ring 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\My Documents\My
Music\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\My
Music\Rhapsody\Bryanbros\Jason Mraz\I''m Yours (Single)\01 - I''m
Yours.wma")
Traceback (most recent call last):
File "<pyshell#10>", line 1, in <module>
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
--
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.
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