Command button open a file(path is in CellA1)
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15 Nov 2011 à 14:11
rizvisa1 Posts 4478 Registration date Thursday 28 January 2010 Status Contributor Last seen 5 May 2022 - 30 Nov 2011 à 17:48
rizvisa1 Posts 4478 Registration date Thursday 28 January 2010 Status Contributor Last seen 5 May 2022 - 30 Nov 2011 à 17:48
Related:
- "Partprograms" "pcdlrn.application"
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1 response
rizvisa1
Posts
4478
Registration date
Thursday 28 January 2010
Status
Contributor
Last seen
5 May 2022
766
30 Nov 2011 à 17:48
30 Nov 2011 à 17:48
If this is working
Set DmisParts = DmisApp.PartPrograms.Open("R\PCDMISW\PartProgram.PRG")
Shell "C:\PCDMISW\PCDLRN.exe C:\PCDMISW\PartProgram.PRG"
then you can change your code as
Set DmisApp = CreateObject("PCDLRN.Application")
Set DmisParts = DmisApp.PartPrograms.Open(ActiveSheet.Cells(1, 1).Value)
Shell "C:\PCDMISW\PCDLRN.exe " & ActiveSheet.Cells(1, 1).Value
or may be this
Shell "C:\PCDMISW\PCDLRN.exe " & ActiveSheet.Cells(1, 1).Value & """"
Set DmisParts = DmisApp.PartPrograms.Open("R\PCDMISW\PartProgram.PRG")
Shell "C:\PCDMISW\PCDLRN.exe C:\PCDMISW\PartProgram.PRG"
then you can change your code as
Set DmisApp = CreateObject("PCDLRN.Application")
Set DmisParts = DmisApp.PartPrograms.Open(ActiveSheet.Cells(1, 1).Value)
Shell "C:\PCDMISW\PCDLRN.exe " & ActiveSheet.Cells(1, 1).Value
or may be this
Shell "C:\PCDMISW\PCDLRN.exe " & ActiveSheet.Cells(1, 1).Value & """"