General information
Category:
Windows API functions
Title:
ShellExecute does not recognize long path names?
I am trying to launch an executable file with the ShellExecute win32api function with the following parameters:
M.SHELL_RET = ShellExecute(FindWindow(0, _SCREEN.caption), ;
"open", "c:\softwares\general\finservices - modelo 2\finservices.exe", ;
"101", "c:\softwares\general\finservices - modelo 2", 1)
The exe does not launch and I receive the following value in M.SHELL_RET: 2
What is happening? I am sure about the exe path and exe file name.
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