Basic Script

[jardineworks]

I'm almost embarassed to ask this but I'm stuck. I have, within my home directory a directory called work and then with in that my company, and with in there a directory for each project.

$HOME/work/mycompany
project-a
project-b
...
project-n

I've created a script (if you can even call it that!) and put it in my $HOME/bin folder. THe idea is that I will be able to type mycompany project-x from anywhere in the terminal and have my directory location changed to the work directory. My script has this right now ---

#!/bin/bash
cd $HOME/work/mycompany

.. with the intention of trapping the arguments once I see this working. I put some echo statements in originally to make sure that it was running, which it is, however, I never change location -- the script ends and I am still where I was when I ran it.

What am I missing -- i know it's something simple but I can't see it.

[sisco311]

It's because the script runs in a subshell.

You can source the script instead of running it as a child:

. yourscript

Or you could simply write a little function or an alias and put it in your ~/.bashrc file.

[CrusaderAD]

Try this:

#!/bin/bash
cd /HOME/work/mycompany

[jardineworks]

It's because the script runs in a subshell.

You can source the script instead of running it as a child:

. yourscript

Or you could simply write a little function or an alias and put it in your ~/.bashrc file.

That worked. Thanks Sisco311.

[AlphaLexman]

Another elegant solution is to add the appropriate pathname to the variable 'CDPATH' in your ~/.bashrc file.

CDPATH="$CDPATH:$HOME/work/mycompany"

And, that way you can simply 'cd' into any project directory!