changing directory

[spuppett]

OK, Im new to UNIX, but so far i am diggin it. I've written a couple of scripts so far, but right now I am stuck on one that shouldn't be hard at all. All i want to do is type in my command and change to the specified directory. My script looks like this:

#!/bin/sh

cd /user/spuppett/Windows_NT/Desktop/LinuxStuff/

#end

when i run it, it does nothing. Whats wrong with it?

 

[danielhozac]

You would have to &quot;source&quot; the script for the parent shell to change to that directory too. For instance, to run it, you'd execute [tt]. /path/to/script[/tt] or [tt]source /path/to/script[/tt] instead of the usual [tt]/path/to/script[/tt] or [tt]<interpreter> /path/to/script[/tt]. //Daniel

 

[spuppett]

sorry man, that didn't help me at all, don't know what you mean. sorry.

Thanks for trying though.

 

[danielhozac]

Here's what happens when you execute your current script.
The shell you are executing it from (the &quot;parent&quot executes whatever program is on the shebang line (in your case [tt]/bin/sh[/tt]) with the script name as a parameter. This creates a new instance of the shell. So when the &quot;child&quot; (the program started by the &quot;parent&quot gets that script name as a parameter, it reads through the file and executes the commands. When it is done, it terminates and all settings are lost. The control is then handed over to the &quot;parent&quot;, which has no idea what the &quot;child&quot; just did, all it knows about it is the exit status.
Now, if you instead &quot;[tt]source[/tt]&quot; the script, the &quot;parent&quot; itself reads the file and executes the command, meaning that the settings and changes are still there when the script ends. This is what you want to do. //Daniel