Extracting part of a file name?

[jchd]

Just to make matters muddier: don't forget that Windows pathes accept both backward and forward slashes!

[Musashi]

This works (at least with backward slashes) :

Local $sString = "|What else (1923).mkv"
ConsoleWrite(StringRegExpReplace ($sString, "^(?=\|)", StringReplace(@ScriptDir, "\", "\\") & "\\") & @CRLF)

@mikell will certainly find a more elegant solution. I am unworthy and also too tired .

Edited December 17, 2020 by Musashi

[AspirinJunkie]

8 hours ago, jchd said:

Just to make matters muddier: don't forget that Windows pathes accept both backward and forward slashes!

For this reason, among others, it is a good idea to clean up the path beforehand with _WinAPI_GetFullPathName() (file does not have to exist here). Then relative paths are evaluated, double backslashes are removed and slashes are converted to backslashes.

[Musashi]

1 hour ago, AspirinJunkie said:

For this reason, among others, it is a good idea to clean up the path beforehand with _WinAPI_GetFullPathName()

Yes, this gives you a consistent notation.

Spoiler

Path            = c:\projects\test
GetFullPathName = c:\projects\test

Path            = c:\\projects\\test
GetFullPathName = c:\projects\test

Path            = c:/projects/test
GetFullPathName = c:\projects\test

Path            = c://projects//test
GetFullPathName = c:\projects\test

Path            = c:\\projects//test
GetFullPathName = c:\projects\test

In this case, e.g. @ScriptDir, the backslashes must be escaped by an additional backslash.

#include <WinAPIFiles.au3>
Local $sString, $sPath
$sString = "|What else (1923).mkv"
; 1. without escaping :
ConsoleWrite(StringRegExpReplace ($sString, "^(?=\|)", @ScriptDir & "\\") & @CRLF)
; 2. with escaping :
$sPath   = StringReplace(_WinAPI_GetFullPathName(@ScriptDir), "\", "\\") & "\\"
ConsoleWrite(StringRegExpReplace ($sString, "^(?=\|)", $sPath) & @CRLF)