mcornbill Posted February 19, 2009 Share Posted February 19, 2009 Hi Guys I'm using the following command to open an Excel file: $FilePath = "c:\testfile.xls" $oExcel = _ExcelBookOpen($FilePath) The Excel file I will be using with this script will be different everytime and will probably not be held in the same location either. So basically rather than declaring the file to open I want it to pop up a windows explorer box, the user browses to the file on their machine, and then that gets read in as $FilePath. I've searched the forums but having difficulty in finding an answer, can anybody help please? Cheers Mark Link to comment Share on other sites More sharing options...
AdmiralAlkex Posted February 19, 2009 Share Posted February 19, 2009 So you searched the forums but not the helpfile? That's not very smart.... Anyway check FileOpenDialog() .Some of my scripts: ShiftER, Codec-Control, Resolution switcher for HTC ShiftSome of my UDFs: SDL UDF, SetDefaultDllDirectories, Converting GDI+ Bitmap/Image to SDL Surface Link to comment Share on other sites More sharing options...
mcornbill Posted February 19, 2009 Author Share Posted February 19, 2009 So you searched the forums but not the helpfile? That's not very smart.... Anyway check FileOpenDialog()Humblest apologies but thanks for your help. Link to comment Share on other sites More sharing options...
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