How to set process priority when starting an executable from a batch file in Windows without launching terminal

Question

I have a .bat file [1] starting from a .vbs script [2] which only launches without launching a terminal if I don't include 'Start /low' in the .bat file.

The 'Start /low' part of the .bat file launches the command with the right (low) priority set, but it launches in a terminal, which I don't want.

I'm only able to launch the desired command in the background without a terminal if I don't set the priority in the .bat file. In which case the final .exe that launches slows my computer down, which is why I want to set its priority to 'low'.

I tried this [3], but it gives me an error when I run it [4].

Would someone kindly tell me how to make the executable start with low priority without launching a terminal window?

[1]

Start /low C:/dataserv-client/dataserv-client.exe --store_path=C:\Users\Chris\StorjData --max_size=800.0GB farm

[2]

Dim WinScriptHost
Set WinScriptHost = CreateObject("WScript.Shell")
WinScriptHost.Run Chr(34) & "C:\Users\Chris\Scripts\start_dataserv-client.bat" & Chr(34), 0
Set WinScriptHost = Nothing

[3]

Dim WinScriptHost
Set WinScriptHost = CreateObject("WScript.Shell")
WinScriptHost.Run Chr(34) & "Start /low C:/dataserv-client/dataserv-client.exe --store_path=C:\Users\Chris\StorjData --max_size=800.0GB farm" & Chr(34), 0
Set WinScriptHost = Nothing

4

Answer 1

The code in [3] didn't work because 1) the closing quote added with chr(34) was in the wrong place - only the executable should be quoted this way, not the entire command line and 2) start isn't a standalone utility which can be executed directly by .Run, it's a command of the command processor cmd (easily checked by running where start in the command prompt console).

CreateObject("WScript.Shell").Run "cmd /c Start /low " & chr(34) & chr(34) & " " & _
     chr(34) & "C:\dataserv-client\dataserv-client.exe" & chr(34) & _
     " --store_path=C:\Users\Chris\StorjData --max_size=800.0GB farm", 0

which executes cmd /c start /low "" "C:\dataserv-client\dataserv-client.exe" ......... - the first "" is for the title parameter of start so that cmd won't confuse it with the quoted exe path.