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Checking Multiple Variables in Bash

Submitted by on September 19, 2017 – 10:04 am

Let’s say you have a dozen variables in your script and you need to check if they have values set. The usual approach can get a bit tedious.

if [[ -z $var1 || -z $var2 || -z $var3 ]]; then

This gets even worse if you need to assign some default value to any variable that is not set:
if [ -z $var1 ]; then var1=unknown; fi
if [ -z $var2 ]; then var2=unknown; fi
if [ -z $var3 ]; then var3=unknown; fi

Here’s a different approach that can save you a lot of typing:
for i in var{1..3}; do
if [ -z "$(eval echo $(echo $`eval echo "${i}"`))" ] ; then
eval "$(echo ${i})"="$(echo "unknown")"
fi; done

If you really have lots and lots of variables that require different default values, you may use a table. Parse this table to populate two arrays: one containing names of the variables and the other – their default values. Let’s say your table.txt looks like this:
variable_two,some other default value
variable_whatever,nothing to see here

We will use it to build two arrays:
array_variables=($(awk -F, '{print $1}' table.txt))
array_defaults=($(awk -F, '{print $2}' table.txt))
unset IFS

Now we can read the array_variables array, check if the variables are set and, if not, assign a corresponding default value from the array_defaults array:
for ((i = 0; i < ${#array_variables[@]}; i++))
	if [ -z "$(eval echo $(echo $`eval echo "${array_variables[$i]}"`))" ]
		eval "$(echo "${array_variables[$i]}")"="$(echo "\"${array_defaults[$i]}\"")"

And to check that the variables have been set:
for i in $(printf '$%s\n' ${array_variables[@]}); do eval echo ${i}; done

some other default value
nothing to see here


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