-----Original Message-----
From: Dan Sommers <[email protected]>
Sent: Friday, February 27, 2026 11:54 AM
To: MRAB <[email protected]>; Em
<[email protected]>
Cc: [email protected]
Subject: Re: Win 10 > Win 11 path error?
On 2026-02-26 at 19:35:13 +0000,
Regarding "Re: Win 10 > Win 11 path error?,"
MRAB <[email protected]> wrote:
> On 26/02/2026 17:53, Em wrote:
> > OutExcelFile = open("_EXCEL-FILE3az.txt","w")
> >
> > The line works in Win 10.
> > In Win 11, if I press F5 it works but if I double click
on the program filename, it fails immediately.
> > No error reported.
> >
> > I believe the problem with the path needing to be
different between Win10 and Win11.
> > Can you tell me what the path should be if that is the
problem?
> >
> > Python: 3.14.1
> >
> "_EXCEL-FILE3az.txt" is a relative path. It looks in the
current
> working directory, wherever that happens to be. It's
better to work
> with full filepaths. The path of the script is given by
`__file__`,
> and its directory by `os.path.dirname(__file__)`, so:
> this_directory =
> os.path.dirname(__file__)
>
> If "_EXCEL-FILE3az.txt" is in the same directory as the
script, its full path will be given by:
>
> data_path =os.path.join(this_directory,
"_EXCEL-FILE3az.txt")
>
> You can then write:
>
> OutExcelFile = open(data_path,"w")
>
> --
>
https://mail.python.org/mailman3//lists/python-list.python.o
rg
On 2026-02-27 at 09:49:11 -0500,
Regarding "Another issue between Win10 and Win 11 using
python,"
Em <[email protected]> wrote:
>
> The statement is: ThisPath = os.getcwd()
>
> For 10 years in Win 10 the line gave me the present path
both when I used F5 in an editor, and when I ran the program
by a double-click on the name of the program in the folder.
>
> I now want to run the program in Win 11.
> As in Win 10, F5 from an editor, it gives me the present
path but when I double-click on the program in the folder,
it gives me some system path on the C drive.
>
> This is now the second line of code that fails like this.
> Can someone explain why.
> Can you suggest a line of code to get the present path for
this that is allowed by Win 11?
I think that's what MRAB already did:
this_directory = os.path.dirname(__file__) data_path =
os.path.join(this_directory, "_EXCEL-FILE3az.txt")
OutExcelFile = open(data_path,"w")
I don't understand that line of code....
--
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rg
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