Python os.path.join 在 Windows

Question

I am trying to learn python and am making a program that will output a script. I want to use os.path.join, but am pretty confused. According to the docs if I say:

os.path.join('c:', 'sourcedir')

I get "C:sourcedir" . According to the docs, this is normal, right?

But when I use the copytree command, Python will output it the desired way, for example:

import shutil
src = os.path.join('c:', 'src')
dst = os.path.join('c:', 'dst')
shutil.copytree(src, dst)

Here is the error code I get:

WindowsError: [Error 3] The system cannot find the path specified: 'C:src/*.*'

If I wrap the os.path.join with os.path.normpath I get the same error.

If this os.path.join can't be used this way, then I am confused as to its purpose.

According to the pages suggested by Stack Overflow, slashes should not be used in join—that is correct, I assume?

Answer 1

To be even more pedantic, the most python doc consistent answer would be:

mypath = os.path.join('c:', os.sep, 'sourcedir')

Since you also need os.sep for the posix root path:

mypath = os.path.join(os.sep, 'usr', 'lib')

Answer 2

Windows has a concept of current directory for each drive. Because of that, "c:sourcedir" means "sourcedir" inside the current C: directory, and you'll need to specify an absolute directory.

Any of these should work and give the same result, but I don't have a Windows VM fired up at the moment to double check:

"c:/sourcedir"
os.path.join("/", "c:", "sourcedir")
os.path.join("c:/", "sourcedir")

Answer 3

To be pedantic, it's probably not good to hardcode either / or \\ as the path separator. Maybe this would be best?

mypath = os.path.join('c:%s' % os.sep, 'sourcedir')

or

mypath = os.path.join('c:' + os.sep, 'sourcedir')

Answer 4

The reason os.path.join('C:', 'src') is not working as you expect is because of something in the documentation that you linked to:

Note that on Windows, since there is a current directory for each drive, os.path.join("c:", "foo") represents a path relative to the current directory on drive C: (c:foo), not c:\\foo.

As ghostdog said, you probably want mypath=os.path.join('c:\\\\', 'sourcedir')

Answer 5

For a system-agnostic solution that works on both Windows and Linux, no matter what the input path, one could use os.path.join(os.sep, rootdir + os.sep, targetdir)

On WIndows:

>>> os.path.join(os.sep, "C:" + os.sep, "Windows")
'C:\\Windows'

On Linux:

>>> os.path.join(os.sep, "usr" + os.sep, "lib")
'/usr/lib'

Answer 6

I'd say this is a (windows)python bug.

Why bug?

I think this statement should be True

os.path.join(*os.path.dirname(os.path.abspath(__file__)).split(os.path.sep))==os.path.dirname(os.path.abspath(__file__))

But it is False on windows machines.

Answer 7

to join a windows path, try

mypath=os.path.join('c:\\', 'sourcedir')

basically, you will need to escape the slash

Answer 8

You have a few possible approaches to treat path on Windows, from the most hardcoded ones (as using raw string literals or escaping backslashes) to the least ones. Here follows a few examples that will work as expected. Use what better fits your needs.

In[1]: from os.path import join, isdir

In[2]: from os import sep

In[3]: isdir(join("c:", "\\", "Users"))
Out[3]: True

In[4]: isdir(join("c:", "/", "Users"))
Out[4]: True

In[5]: isdir(join("c:", sep, "Users"))
Out[5]: True

Answer 9

Consent with @georg-

I would say then why we need lame os.path.join - better to use str.join or unicode.join eg

sys.path.append('{0}'.join(os.path.dirname(__file__).split(os.path.sep)[0:-1]).format(os.path.sep))

Answer 10

answering to your comment : "the others '//' 'c:', 'c:\\\\' did not work (C:\\\\ created two backslashes, C:\\ didn't work at all)"

On windows using os.path.join('c:', 'sourcedir') will automatically add two backslashes \\\\ in front of sourcedir .

To resolve the path, as python works on windows also with forward slashes -> '/' , simply add .replace('\\\\','/') with os.path.join as below:-

os.path.join('c:\\\\', 'sourcedir').replace('\\\\','/')

eg: os.path.join('c:\\\\', 'temp').replace('\\\\','/')

output : 'C:/temp'

Answer 11

The proposed solutions are interesting and offer a good reference, however they are only partially satisfying. It is ok to manually add the separator when you have a single specific case or you know the format of the input string, but there can be cases where you want to do it programmatically on generic inputs.

With a bit of experimenting, I believe the criteria is that the path delimiter is not added if the first segment is a drive letter, meaning a single letter followed by a colon, no matter if it corresponds to a real unit.

For example:

import os
testval = ['c:','c:\\','d:','j:','jr:','data:']

for t in testval:
    print ('test value: ',t,', join to "folder"',os.path.join(t,'folder'))

 test value: c: , join to "folder" c:folder test value: c:\\ , join to "folder" c:\\folder test value: d: , join to "folder" d:folder test value: j: , join to "folder" j:folder test value: jr: , join to "folder" jr:\\folder test value: data: , join to "folder" data:\\folder

A convenient way to test for the criteria and apply a path correction can be to use os.path.splitdrive comparing the first returned element to the test value, like t+os.path.sep if os.path.splitdrive(t)[0]==t else t .

Test:

for t in testval:
    corrected = t+os.path.sep if os.path.splitdrive(t)[0]==t else t
    print ('original: %s\tcorrected: %s'%(t,corrected),' join corrected->',os.path.join(corrected,'folder'))

 original: c: corrected: c:\\ join corrected-> c:\\folder original: c:\\ corrected: c:\\ join corrected-> c:\\folder original: d: corrected: d:\\ join corrected-> d:\\folder original: j: corrected: j:\\ join corrected-> j:\\folder original: jr: corrected: jr: join corrected-> jr:\\folder original: data: corrected: data: join corrected-> data:\\folder

it can be probably be improved to be more robust for trailing spaces, and I have tested it only on windows, but I hope it gives an idea. See also Os.path : can you explain this behavior? for interesting details on systems other then windows.

Answer 12

I got around this by using:

os.sep.join(list('C:', 'sourcedir'))

join here is not from os.path.join() , but from ''.join()

It can be useful in the following situation:

import os

some_path = r'C:\some_folder\some_file.txt'
path_items = some_path.split(os.sep)
same_path = os.sep.join(path_items)

Answer 13

For simplicity's sake, a workaround:

my_path = r'D:\test\test2\file.txt' # original path string
drive_letter = my_path.split(':')[0] # avoid os.path.join relative path bs
my_path_parts = os.path.normpath(my_path.split(':')[1]).split(os.sep)

# do some path cmponent modifications here if you want

if drive_letter: # if drive letter exists
    drive_letter += ':\\'

my_new_path = drive_letter + os.path.join(*my_path_parts)
my_new_path

Answer 14

In windows using os.paht.join("/", "Temp") will result in /Temp by default but as strange as it sounds, there is no problem in using that path as a full path equivalent to "C:/Temp" and it works for both saving files and opening files.