Jacquesvdz
Programmer
If I have a Filepath
strFilePath = C:\HORIZON DEV\HorizonPackage\Development\Horizon\HORIZON Debug.VBG
How do I get the FilePath like this: C:\HORIZON DEV\HorizonPackage\Development\Horizon\..
without the FileName.vbg
and get the FileName alone
strFileName = HORIZON Debug.VBG
So on the end it I must get this
StrFilePath = C:\HORIZON DEV\HorizonPackage\Development\Horizon
strFileName = HORIZON Debug.VBG
from the original
strFilePath = C:\HORIZON DEV\HorizonPackage\Development\Horizon\HORIZON Debug.VBG
Keep in mind the URL will always differ.So it has to be the Filename after the last "\"
Any Idea how to just split fileName from the path
strFilePath = C:\HORIZON DEV\HorizonPackage\Development\Horizon\HORIZON Debug.VBG
How do I get the FilePath like this: C:\HORIZON DEV\HorizonPackage\Development\Horizon\..
without the FileName.vbg
and get the FileName alone
strFileName = HORIZON Debug.VBG
So on the end it I must get this
StrFilePath = C:\HORIZON DEV\HorizonPackage\Development\Horizon
strFileName = HORIZON Debug.VBG
from the original
strFilePath = C:\HORIZON DEV\HorizonPackage\Development\Horizon\HORIZON Debug.VBG
Keep in mind the URL will always differ.So it has to be the Filename after the last "\"
Any Idea how to just split fileName from the path