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[Georgy Pruss]
> On Windows XP glob.glob doesn't work properly for files without
> extensions.

I'd say it's Microsoft's dir that doesn't work properly in this case.

> E.g. C:\Temp contains 4 files: 2 with extensions, 2
> without.
>
> C:\Temp>dir /b *
> aaaaa.aaa
> bbbbb.bbb
> ccccc
> ddddd
>
> C:\Temp>dir /b *.
> ccccc
> ddddd

Why on Earth should a pattern specifying a period match filenames that don't
contain a period? Would you expect

*x

to match

abc

? Nope. It's dir that special-cases the snot out of a period, not Python's
glob. For cross-platform sanity, glob has to work the same way across
platforms, and the Unixish shells have the obvious, explainable,
unsurprising semantics here:

$ ls *
aaaaa.aaa bbbbb.bbb ccccc ddddd

$ ls *.
ls: *.: No such file or directory

$ ls *.*
aaaaa.aaa bbbbb.bbb

> C:\Temp>python
> Python 2.3 (#46, Jul 29 2003, 18:54:32) [MSC v.1200 32 bit (Intel)]
> on win32 Type "help", "copyright", "credits" or "license" for more
> information.
> >>> import glob
>
> >>> glob.glob( '*' )
> ['aaaaa.aaa', 'bbbbb.bbb', 'ccccc', 'ddddd']
>
> >>> glob.glob( '*.' )
> []

> It looks like a bug.

Good luck getting Microsoft to fix it <wink>.

"Tim Peters" <> wrote in
news:mailman.1194.1070165783.702.python-:

> ? Nope. It's dir that special-cases the snot out of a period, not
> Python's glob.

On that basis Python also 'special-cases the snot out of a period': try
win32api.FindFiles('*.') to get the same behaviour as dir.

More accurate would be to say that the operating system has consistent but
weird behaviour, since it is below the application level that trailing dots
are stripped from filenames and ignored from file patterns. The only way to
avoid it on Windows is to retrieve all the filenames and then perform your
own pattern matching on them.

glob.glob is bypassing the operating system's filename matching to provide
behaviour which is consistent across different operating systems, but
inevitably this means that it doesn't match some platform's standard
behaviour.

--
Duncan Booth
int month(char *p){return(124864/((p[0]+p[1]-p[2]&0x1f)+1)%12)["\5\x8\3"
"\6\7\xb\1\x9\xa\2\0\4"];} // Who said my code was obscure?