Running a program after copy file, in one Windows CMD line?

Question

I try to create visual studio post-build event command line and i want:

to kill process if exist;
copy file;
launch a program;

start taskkill /f /im app.exe /t & xcopy /y "$(TargetPath)" "C:\Program Files (x86)\path\$(TargetFilename)" & start "" "C:\Program Files (x86)\program\app.exe"

I use start taskkill /f /im app.exe /tinstead of taskkill /f /im app.exe /t because if the process not exists i get not found error

Now all tree command runs OK with no error but the last command not launch the program but if i execute this command separately the program is executed.

How can i fix this?

Do you need to pass that first empty string argument to the third `start`? — An0nC0d3r, Dec 26 '15 at 17:02
here i see the first argument is the title http://stackoverflow.com/a/17298306/2327332 — TotPeRo, Dec 26 '15 at 17:07

score 1 · Accepted Answer · answered Dec 26 '15 at 17:44

1

You have a Timing Problem.

Try:

start /wait taskkill /f /im app.exe /t & xcopy /y "D:\Workspaces\Dev\path\bin\Debug\my.dll" "C:\Program Files (x86)\path\my.dll" & start "" "C:\Program Files (x86)\program\app.exe"

answered Dec 26 '15 at 17:44

Stephan

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yes. this was the problem. this ```wait ``` solve my problem.thanks Stephan. – TotPeRo Dec 26 '15 at 18:10

Running a program after copy file, in one Windows CMD line?

1 Answers1