Declaring variables in external script

Question

I'm tying to write a script which uses variables declared in another script. I've tried so far using export to define environmental variables or just using local ones.

So I have my config script:

export DIR_STORAGE="/var/storage/"

And script trying to use it in the same directory:

source conf.sh
echo "DIR="${DIR_STORAGE}

but that doesn't seem to work. How can I fix this?

don not forget to include shebangs in the beginning of your scirpts: #!/bin/bash -> en.wikipedia.org/wiki/Shebang_(Unix) Also try to specify full path to conf.sh (for testing). — hovanessyan
– hovanessyan, Commented Jul 23, 2012 at 9:25
Hm, it worked with absolute path, thanks. Although this doesn't look like an elegant solution. — Disenchanted
– Disenchanted, Commented Jul 23, 2012 at 9:43

Costi Ciudatu · Accepted Answer · 2012-07-27 16:58:03Z

5

Even if you have both files in the same directory, source conf.sh will look for conf.sh in the current directory. For example:

SHELL$ pwd
/home/user

SHELL$ ls mydir/
script.sh conf.sh

SHELL$ mydir/script.sh
script.sh: line 1: conf.sh: No such file or directory

In this case, the script will look for 'conf.sh' in the current (home) directory, not under mydir.

To avoid this and always look for the config file in the same directory where the script is, a common practice is to use:

source "$(dirname $0)/conf.sh"

This will extract the path from the command you used to invoke the script and append the config file name to it; in the example above, this will become source "mydir/conf.sh" which should work fine.

P.S. If you use source there's no need to export any variables.

answered Jul 27, 2012 at 16:58

Costi Ciudatu

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