cheeseandcereal Posted July 13, 2012 Posted July 13, 2012 (edited) Hey so I was wondering if I had a file string such as "C:/example/file.exe" would there be any way to pull out the directory from that so I could set a variable to be "C:/example/" only knowing "C:/example/file.exe" automatically? So for example, If I had this script: Opt("MustDeclareVars", 1) Global $file1 Global $directory1 $file1 = FileOpenDialog ("", "", "All Files (*.*)") how would I be able to set $directory1 as the directory containing selected $file1 Edited July 13, 2012 by cheeseandcereal
water Posted July 13, 2012 Posted July 13, 2012 You could use function _PathSplit ore use a regular expression (some examples can be found on the forum). My UDFs and Tutorials: Spoiler UDFs: Active Directory (NEW 2024-07-28 - Version 1.6.3.0) - Download - General Help & Support - Example Scripts - Wiki ExcelChart (2017-07-21 - Version 0.4.0.1) - Download - General Help & Support - Example Scripts OutlookEX (2021-11-16 - Version 1.7.0.0) - Download - General Help & Support - Example Scripts - Wiki OutlookEX_GUI (2021-04-13 - Version 1.4.0.0) - Download Outlook Tools (2019-07-22 - Version 0.6.0.0) - Download - General Help & Support - Wiki PowerPoint (2021-08-31 - Version 1.5.0.0) - Download - General Help & Support - Example Scripts - Wiki Task Scheduler (2022-07-28 - Version 1.6.0.1) - Download - General Help & Support - Wiki Standard UDFs: Excel - Example Scripts - Wiki Word - Wiki Tutorials: ADO - Wiki WebDriver - Wiki
Spiff59 Posted July 13, 2012 Posted July 13, 2012 The DIY method would be something like: $path = StringLeft($file, StringInStr($file, "/", 0, -1))
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