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I have created a executable file lets say 'Startbackup.exe' and it is located in 'D:\Start Backup\Startbackup.exe'

What i am trying to acheive is to create a bactch file to start this .Exe uisng defualt paramter . My parameter is 'Startbackup/run'.

I tried following command in cmd and it works.

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How can i create a .bat file? I have tried following but its not working

Start /D D:\Start Backup\ Startbackup Parameter Startbackup/run
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  • This may help: stackoverflow.com/questions/72671/… Commented Jun 29, 2018 at 13:44
  • @RuudHelderman thanks for the link it helps Commented Jun 29, 2018 at 13:57
  • Your 'starting directory path' has spaces so should be doublequoted, Start /D "D:\Start Backup". If startbackup/run works then it appears that /run is your parameter, so all you need is to append that, with an optional/empty title, e.g Start /D "D:\Start Backup" startbackup /run or Start /D "D:\Start Backup" "" startbackup /run. Commented Jun 29, 2018 at 14:55

2 Answers 2

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That is because the first parameter of start is the name, and the second one is the path to the executable.

Try this:

Start "" "D:\Start Backup\StartBackup.exe"
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I got it working now.

start "" "D:\Start Backup\StartBackup.exe" parameter StartBackup/Start

I dont have any space in my arguments so this solution works.

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