CentOS 7 -eq: unary operator expected

Question

I have to check Tomcat 8 is running or not. For this I am using the below script.

#!/bin/bash

statuscode=$(wget --server-response http://localhost:8080 2>&1 | awk '/^  HTTP/{print $2}')

if [ $statuscode -eq 200 ]
then
    echo "TOMCAT OK"
    exit 0
else
    echo "TOMCAT CRITICAL"
    exit 2
fi

When I run this script on CentOS 7.

If Tomcat 8 is running then script is running without any error.

If Tomcat 8 is stopped then script is running with following error

line 5: [: -eq: unary operator expected

How can I fix this issue?

This is because `$statuscode` get populated with an empty string or something strange. Try to debug by `echo`ing the variable before the `if` condition. — fedorqui, Aug 09 '16 at 10:45
@fedorqui yes when tomcat stopped $statuscode is empty. How can i handle this? — user3441151, Aug 09 '16 at 10:47
If you were quoting correctly (`[ "$statuscode" -eq 200 ]`), you'd be getting a more useful error; it's the lack of quotes that causes the shell to string-split your variable into an unknown number of arguments (in this case 0) before passing those arguments to the `[` command (aka test). — Charles Duffy, Aug 09 '16 at 13:02
The canonical may be *["unary operator expected" error in Bash if condition](https://stackoverflow.com/questions/13617843/)* — Peter Mortensen, Oct 05 '21 at 16:56

score 2 · Accepted Answer · answered Aug 09 '16 at 10:53

2

Check if the variable is not empty before comparing it against the expected output.

#!/bin/bash

statuscode=$(wget --server-response http://localhost:8080 2>&1 | awk '/^  HTTP/{print $2}')

if [ -n "$statuscode" ] && [ $statuscode -eq 200 ]
then
    echo "TOMCAT OK"
    exit 0
else
    echo "TOMCAT CRITICAL"
    exit 2
fi

answered Aug 09 '16 at 10:53

Thirupathi Thangavel

2,418
3
29
49

If my tomcat 8 service is running then "wget --server-response http://localhost:8080 2>&1 | awk '/^ HTTP/{print $2}'" would give me always 200 or something else value? – user3441151 Aug 09 '16 at 11:21
There should be better way of checking if a service is running or not. Like a `systemctl` command – Thirupathi Thangavel Aug 09 '16 at 11:32

score 0 · Answer 2 · edited Oct 05 '21 at 16:58

0

Try this:

If statuscode is empty, then throw -eq: unary operator expected.

#!/bin/bash
{
statuscode=$(wget --server-response http://localhost:80 2>&1 | awk '/^  HTTP/{print $2}')

if [ -z "$statuscode" ]
then
echo "TOMCAT CRITICAL";
exit 2;
else
  if [ $statuscode -eq 200 ]
    then
     echo "TOMCAT OK";
     exit 0;
  fi
fi
}

edited Oct 05 '21 at 16:58

Peter Mortensen

30,738
21
105
131

answered Aug 09 '16 at 10:54

Mustafa DOGRU

3,994
1
16
24

Re *"then throw -eq: unary operator expected."*: Where is that happening? Can you elaborate? E.g, what do you mean by ***"throw"***? Please respond by [editing (changing) your answer](https://stackoverflow.com/posts/38848796/edit), not here in comments (***without*** "Edit:", "Update:", or similar - the answer should appear as if it was written today). – Peter Mortensen Oct 05 '21 at 17:00

CentOS 7 -eq: unary operator expected

2 Answers2