-eq: unary operator expected, shell argument parsing

Question

I am getting this error while comparing shell argument to a string. If the code look like this:

online=true
if [ "$2" -eq '-o' ]
then
    online=false
fi
echo $online

Then for e.g. input I am getting those results:

$ ./currency.sh 2 -o
./currency.sh: line 13: [: -o: integer expression expected
true

-eq is an arithmetic test operator - for string comparison, use = — steeldriver, Mar 21 '18 at 01:56
Thank you. I'm not used to bash, Ruby is all I need for shell usually :) — siery, Mar 21 '18 at 01:59
Possible duplicate of Conditional statement, "unary operator expected" — Sergiy Kolodyazhnyy, Mar 21 '18 at 05:26

Vlastimil Burián · Accepted Answer · 2018-03-21T12:57:35.773

4

In this line:

if [ "$2" -eq '-o' ]

You have used arithmetic operator -eq which takes the second argument as a number.

Naturally, it fails for that reason alone.

When comparing strings, you can use POSIX `=` operator:

if [ "$2" = '-o' ]

Note, that this version should work in all shells, as it is defined by POSIX (Portable Operating System Interface).

if [[ "$2" == '-o' ]]

Note, that this version will work in only Bash (Bourne-again shell) and alike.

In contrast, double brackets [[ .. ]] and == operator are both defined in Bash only and will not work in other shells.

edited Mar 21 '18 at 12:57

answered Mar 21 '18 at 05:05

Vlastimil Burián

Thank you! Cold you explain in detail, why do you use double '[' around statement? I have tried both singular and double version of this and both works. – siery Mar 21 '18 at 12:24