I'm terribly new to scripting on windows. Using windows 7 64.
I'm trying to make a .bat file that I can double click, and have it open a command prompt and automatically cd
me to a certain directory.
I tried making a .bat file with
@ECHO OFF
cmd "cd C:\my\destination"
Which opens what looks like a command prompt, but doesn't seem to let me type any commands.
I then tried:
@ECHO OFF
start cmd "cd C:\my\destination"
But this just sent me into a loop opening tons and tons of prompts until my computer crashed :) The .bat file was located in the destination directory if that matters.
The /D will change folder and drive and works on 2000+ (Not sure about NT4)
If you take a look at Vista's open command here, it uses
cmd.exe /s /k pushd \"%V\"
but I don't think %V is documented. Using pushd is a good idea if your path is UNC (\\server\share\folder
) To get UNC current directory working, you might have to set the DisableUNCCheck registry entry...This works for me:
The quoted string is actually two commands (separated by a double ampersand): The first command is to change to the specified directory, the second command is to change to the specified drive letter.
Put this in a batch (.BAT) file and when you execute it you should see a Command Prompt window at the specified directory.
just open a text editor and type
Then save it as a .bat file. Works for me.
Why so complicated? Just create an alias to cmd.exe, right click on the alias and navigate to its settings. Change the "execute in" to the path you want to have as standard path. It will always start in this path.
Use the /K switch:
But IMHO, the most useful switch is /?.
And only if it does not work, then Google it, as @Neeraj suggested :D
This could be done like that:
If you need to execute a file or command after you open the cmd you can just replace the last line with: