file names longer than MAX_PATH under Windows 2003

I try to open file with pathname length 282 bytes:
E:\files\..................\something.dat

On MSDN (http://msdn.microsoft.com/library/default.asp?url=/library/en-us/fileio/fs/naming_a_file.asp) described method to access
files with path length
up to 32000 bytes: just add prefix \\?\ to file name.
But when I try to pass prefixed name to file(), I get the same result as when I don't add the prefix: file not found. May be Python
just doesn't support long unicode filenames?
Is there way to open such files?

| [... MS advise ...] just add prefix \\?\ to file name.

With a file called c:\temp\test.txt, I successfully
opened and read it like this:

print open (r"\\?\C:\temp\test.txt").read ()

So the basic functionality works. I didn't
artificially generate a long path to see if
there's a problem in that direction.

But note that r prefix to the string. Is it possible
that your string didn't include it? If not, then the
backslash character which Windows uses as a separator
can be stolen by Python which sees it as an escaping
character.

TJG

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Backslashes have special meaning to Python and need to be escaped. If you
do this:

f = file("E:\files\...\something.dat", "r")

Python's string escape rules means you are actually trying to open the
file "E:files...something.dat" which doesn't exist.

You should escape the backslashes:

f = file("E:\\files\\...\\something.dat", "r")

or use raw strings:

f = file(r"E:\files\...\something.dat", "r")

or just use forward slashes and let Windows deal with it:

f = file("E:/files/.../something.dat", "r")

Does this help?

--
Steven.

It's ok with backslashes, I don't use r'', but double them so string is correct:
>>> print c
\\?\e:\files\........

Not to state the obvious, but can you cut-and-paste that long
string (the one starting with \\?\e:\...) from the Python
interpreter into the [S]tart [R]un [O]pen field to see what
comes up? I'm just trying to make sure of the most straightforward
fact: that the file you've got there definitely does exist!

backslashes quoted.

>>> print c
\\.\e:\files\xxxxxxxxxxxxxxxx\xxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxx\xxxxxxxxxxx xxxxxxxxx
\xxxxxxxxxxxxxxxxxxxxxxx\xxxxxxxxxxxxxx\xxxxxxxxxxxxxxxxxxxxx\xxxxxxxxxxxx
xxxxxxxxxxxxxxx\xxxxxxxxxxxxxxxxxxxxxxxxx\xxxxxxxxxxxxxxxxxxxxxxx\xxxxxx xxxxxxxxxx\xxxxxxxxxxxxxxxxxxxxxxxxxxxx.xls

(cyrillic letters in filename here I replaced with x-es)

>>> file(c,"rb")
Traceback (most recent call last):
File "<stdin>", line 1, in ?
IOError: [Errno 2] No such file or directory: u'\\\\.\\e:\\files\\\u041f\u0420....filename skipped....\u0442.xls'

>>> win32file.CreateFile(c,win32file.GENERIC_READ,0,None,win32file.OPEN_EXISTING
,0,None)
Traceback (most recent call last):
File "<stdin>", line 1, in ?
pywintypes.error: (206, 'CreateFile', 'The filename or extension is too long.')

When I open explorer, I can browse folder and see this file (but I can do almost nothing with it). It was created through share.
Ntbackup can work with the file - why Python can't?

[slaps head]
I seem to be a bit confused about string escaping rules. Only some
backslashes have special meaning, the rest remain in the string.

Sergey, you said UNICODE file names. Not just ordinary strings.

Are you passing a unicode object to the function?

f = file(u"E:\\files\\...\\something.dat", "r")

--
Steven.

I cannot do this: when I paste filename there, trail of filename is missing due to length limit in input line.
But I strongly suppose that file exists as its name was get through os.listdir.

I pass variable c into functions:

>>> c
u'\\\\.\\e:\\files\\\u041f\u0420\u041e\u0414\u041e \u041c\u0435\u043d\u0435\u043
[many unicode chars skipped]
44b\u0439_\u0411\u044e\u0434\u0436\u0435\u0442.xls'

Sorry, it was my error - there was extra backslash in the middle of path.
\\?\path works OK.

I see from another post that CreateFile cannot open your file.
That puts it further away from Python, although it doesn't
explain how some other program can see the files. Can you use
os.startfile (or its equivalent win32api.ShellExecute from
pywin32)? Perhaps if you were to chdir to the directory in
question you'd be able to access the file.

Don't know really; clutching at straws.

I found error in filename (extra backslash) and now I can open file.
But another trouble appeared: I cannot listdir so long name.
I examined posixmodule.c... There are wcsncpy with hard limit of MAX_PATH*2+5.
So... RIP, my module...

Claudio

Have a look at win32file.FindFilesIterator from the pywin32 extensions.
Maybe that can cope? (I haven't looked at the source).

<code>
import win32file
for (
attr, created, accessed, written, size_hi, size_lo,
_, _,
filename, alt_filename
) in win32file.FindFilesIterator ("c:/temp/*"):
print filename

</code>

Yeah, it works!
THANK YOU!
(but now I must have two pieces of code, one for linux and one for windows)
Interesting, why developers of python didn't use here all power of win32 API?

It seems this method doesn't help here.

It will in Python 2.5.

Regards,
Martin