FileOpenDialog problem

[MasterX]

If u use the FileOpenDialog function, you get the path of the file like C:\Program Files\somedir\somefile.exe

I would like to get the path without the file C:\Program Files\somedir\ , so i used stringsplit with "\" as parameter, but i don´t know how to make the arrays in one variable.

Would you help me please

[FireFox]

@MasterX

$fod = FileOpenDialog(your param)
$file = StringSplit($fod, '\')
MsgBox(64, 'File', $file[$file[0]])

Cheers, FireFox.

[MasterX]

Thanks champ

[BrettF]

_PathSplit() is also another option

Cheers,

Brett

[MrCreatoR]

RegExp as well:

$sFilePath = FileOpenDialog("Open file...", "", "All (*.*)")

$sFileName = StringRegExpReplace($sFilePath, "^.*\\", "")
MsgBox(64, "File Name", $sFileName)

[trancexx]

RegExp as well:

$sFilePath = FileOpenDialog("Open file...", "", "All (*.*)")

$sFileName = StringRegExpReplace($sFilePath, "^.*\\", "")
MsgBox(64, "File Name", $sFileName)

I find this solution to be most compact and used it many times.

...only, what is the meaning of circumflex accent (^) there?

[MrCreatoR]

I find this solution to be most compact and used it many times.

...only, what is the meaning of circumflex accent (^) there?

As far as i remember, it's indicating a begining of string (equivalent to \A). I think it's can be ommited in this case.