Evaluating variables from a list of names in bash

Question

vars="a,b"
a="True"
b="False"
IFS=","
for var in $vars; do
  if [[ "$var" = "True" ]]; then
    echo "True found"    
  fi
done

I would expect the above bash script to print out "True found". But it does not print anything. Any ideas as to why ?

You want indirect variables. See mywiki.wooledge.org/BashFAQ/006 — Charles Duffy
– Charles Duffy, Commented Jun 15, 2015 at 23:32
You are expecting bash to double-evaluate the variable contents. It isn't going to do that. You need to do the second level of indirection yourself. — Etan Reisner
– Etan Reisner, Commented Jun 15, 2015 at 23:33
...as it is, err, why would you expect $var to evaluate to True or False, rather than a or b? — Charles Duffy
– Charles Duffy, Commented Jun 15, 2015 at 23:33
Add set -x to the top of this script to see what is actually happening at each step to better understand what is going on. — Etan Reisner
– Etan Reisner, Commented Jun 15, 2015 at 23:34

Charles Duffy · Accepted Answer · 2015-06-15 23:34:25Z

2

Make this:

if [[ "${!var}" = "True" ]]; then

${!varname} expands the variable named in $varname. Otherwise, you get the name itself, not the contents of the variable with that name.

See BashFAQ #6 for far more details.

answered Jun 15, 2015 at 23:34

Charles Duffy

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1 Comment

Indirect variables. Nice. TIL.