Windows文件路径的Python正则表达式

Question

The problem, and it may not be easily solved with a regex, is that I want to be able to extract a Windows file path from an arbitrary string. The closest that I have been able to come (I've tried a bunch of others) is using the following regex:

[a-zA-Z]:\\([a-zA-Z0-9() ]*\\)*\w*.*\w*

Which picks up the start of the file and is designed to look at patterns (after the initial drive letter) of strings followed by a backslash and ending with a file name, optional dot, and optional extension.

The difficulty is what happens, next. Since the maximum path length is 260 characters, I only need to count 260 characters beyond the start. But since spaces (and other characters) are allowed in file names I would need to make sure that there are no additional backslashes that could indicate that the prior characters are the name of a folder and that what follows isn't the file name, itself.

I am pretty certain that there isn't a perfect solition (the perfect being the enemy of the good) but I wondered if anyone could suggest a "best possible" solution?

Answer 1

Here's the expression I got, based on yours, that allow me to get the path on windows : [a-zA-Z]:\\\\((?:[a-zA-Z0-9() ]*\\\\)*).* . An example of it being used is available here : https://regex101.com/r/SXUlVX/1

First, I changed the capture group from ([a-zA-Z0-9() ]*\\\\)* to ((?:[a-zA-Z0-9() ]*\\\\)*) .
Your original expression captures each XXX\\ one after another (eg : Users\\ the Users\\ ).
Mine matches (?:[a-zA-Z0-9() ]*\\\\)* . This allows me to capture the concatenation of XXX\\YYYY\\ZZZ\\ before capturing. As such, it allows me to get the full path.

The second change I made is related to the filename : I'll just match any group of character that does not contain \\ (the capture group being greedy). This allows me to take care of strange file names.

Another regex that would work would be : [a-zA-Z]:\\\\((?:.*?\\\\)*).* as shown in this example : https://regex101.com/r/SXUlVX/2

This time, I used .*?\\\\ to match the XXX\\ parts of the path.
.*? will match in a non-greedy way : thus, .*?\\\\ will match the bare minimum of text followed by a back-slash.

Do not hesitate if you have any question regarding the expressions.
I'd also encourage you to try to see how well your expression works using : https://regex101.com . This also has a list of the different tokens you can use in your regex.

Edit : As my previous answer did not work (though I'll need to spend some times to find out exactly why), I looked for another way to do what you want. And I managed to do so using string splitting and joining.
The command is "\\\\".join(TARGETSTRING.split("\\\\")[1:-1]) .
How does this work : Is plit the original string into a list of substrings, based. I then remove the first and last part ( [1:-1] from 2nd element to the one before the last) and transform the resulting list back into a string.

This works, whether the value given is a path or the full address of a file. Program Files (x86)\\\\Adobe\\\\Acrobat Distiller\\\\acrbd.exe fred is a file path Program Files (x86)\\\\Adobe\\\\Acrobat Distiller\\\\acrbd.exe fred\\ is a directory path