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up vote 7 down vote favorite

I'd like to make sure that at a certain point of a script, after sourceing a configuration file, several variables are set and, if they are not, to stop execution, telling the user about the missing variable. I have tried

for var in $one $two $three ; do
    ...

but if for example $two is not set, the loop is never executed for $two. The next thing I tried was 

for var in one two three ; do
    if [ -n ${!var} ] ; then
        echo "$var is set to ${!var}"
    else
        echo "$var is not set"
    fi
done

But if two is not set, I still get "two is set to" instead of "two is not set".

How can I make sure that all required variables are set?

Update/Solution: I know that there is a difference between "set" and "set, but empty". I now use (thanks to http://stackoverflow.com/a/16753536/3456281 and the answers to this question) the following:

if [ -n "${!var:-}" ] ; then

so, if var is set but empty, it is still considered invalid.

share|improve this question
1  
You can also add set -u to the beginning of your script to terminate it immediately when an unset variable is used. –  n.st 11 hours ago
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4 Answers

up vote 5 down vote accepted

Quoting error.

if [ -n "${!var}" ] ; then

For the future: Setting

set -x

before running the code would have shown you the problem. Instead of adding that to the code you can call your script with

bash -vx ./my/script.sh
share|improve this answer
    
This works, but what is happening with/whithout quotes? Is my general approach correct the first place? –  Jasper 12 hours ago
    
Yeah, that's precognitive: I answered the question before it was asked... 8-) –  Hauke Laging 12 hours ago
2  
@Jasper You should always quote variables. That does not cost more time than thinking about whether this is necessary every time. But it avoids the errors. –  Hauke Laging 12 hours ago
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up vote 4 down vote

Only thing you need are quotes in your test:

for var in one two three ; do
    if [ -n "${!var}" ] ; then
        echo "$var is set to ${!var}"
    else
        echo "$var is not set"
    fi
done

Works for me.

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up vote 1 down vote

I think if you mean not set, so the variable must never be initialized. If you use [ -n "${!var}" ], so the empty variable like two="" will be failed, while it is set. You can try this:

one=1
three=3

for var in one two three
do
    declare -p $var > /dev/null 2>&1 \
    && echo $var is set to ${!var} \
    || echo $var is not set
done
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up vote 1 down vote

You can add

set -u

to the beginning of your script to make it terminate when it tries to use an unset variable.

A script like

#!/bin/sh
set -u
echo $foo

will result in

script.sh: 3: script.sh: foo: parameter not set

If you're using bash instead, the error will look like this:

script.sh: line 3: foo: unbound variable

share|improve this answer
    
And in your opinion this makes which sense for run time checks? –  Hauke Laging 9 hours ago
    
@HaukeLaging I don't quite follow you — set -u prevents exactly the kind of error the OP is trying to avoid and (in contrary to all other solutions) isn't limited to a specific set of variables. It is in fact a useful precaution for almost all shell scripts to have them fail safely instead of doing unexpected things when a variable is unset. This article is a handy reference on the topic. –  n.st 9 hours ago
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