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Post by honky on Jan 7, 2024 12:16:18 GMT
Hello everyone, In a folder, I have a: "name.exe" file. In the same folder, I have a folder: "launcher" in which there is a "launcher.bas". I would like to launch: "name.exe" since "launcher.bas" In: "Launcher.bas", I tried: run "name.exe.." run "name.exe..\" run "name.exe\.." Nothing works, how to do it? Thank you for...
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Post by Rod on Jan 7, 2024 12:52:58 GMT
Read that again try run "/name,exe" you need to give the path to where name.exe is. so try \name.exe or /name.exe
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Post by xxgeek on Jan 7, 2024 13:28:27 GMT
The following works for me.
Run "..\name.exe"
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Post by honky on Jan 7, 2024 13:45:39 GMT
Thank you xxgeek,it's work.
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Post by Rod on Jan 8, 2024 10:26:21 GMT
A little more flesh on the bones.
Every directory has an implicit '.' directory that refers to itself, and an implicit '..' directory that refers to its parent.
so
./ is the current directory
../ is up one directory
../../ is up two directories
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Post by honky on Jan 9, 2024 13:03:09 GMT
Yes, "filedialog "Open", "..\*.exe", fileName$" is good "filedialog "Open", "*.exe..\", fileName$" is not good"
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