Windows – Run a file with HIGH priority

batch filecommand linewindows 10

I understand how to run a process or program with HIGH priority, but I want to run an actual FILE (which of course is opened with a program) in high priority using command line in a batch file.

For example, if I wanted to open a program (.exe file) in high priority, I would do:

start "" /HIGH c:\windows\system32\msiexec.exe

I can easily do this in a batch file.
But what if I wanted to open a particular file, like:

c:\user\username\desktop\file.msi

I've tried:

start "" /HIGH c:\user\username\desktop\file.msi

but it doesn't open in high priority. Only when calling a program, and not a specific file that is opened with a program does it run with high priority.

How can I run file.msi in high priority from my batch file?

Best Answer

  • For any file that requires a handler or runtime to run, you run the handler or runtime application as /high and then use whichever method is appropriate for that application to call the file you wish it to run.

    As EBGreen notes, for msi files you run msiexec with high priority, and then pass the installer argument and path.

    start /high "msiexec /i installer.msi"