[SOLVED] If condition is throwing error

vishnukumar · 11-01-2010, 06:42 AM

Hi,

I have created a shell script to find, whether the process is running or not. Below is my shell script.

Code:

#!/bin/bash

result=$(ps -ef | grep -v grep | grep "applicationname" | awk '{print $2}')
if [$result == ""];
then
echo "process is not running"
else
echo "process is running"
fi

When I execute this script, while the process (applicationname) is running, I get an error message like

"line 5: [8017: command not found
process is running"

But, I can say that "8017" is my process id. I don't know, how to fix this error? Why it throws?

Thanks
-Vishnu

druuna · 11-01-2010, 07:09 AM

Hi,

Mind the space(s) between [ and $result and " and ].

if [ $result == "" ];

Hope this helps.

vishnukumar · 11-01-2010, 08:09 AM

Thanks druuna, it works fine. But, after the fix I got another error "line 5: [: ==: unary operator expected", then I google it and found that I have to add double quotes to the $result to fix it. Below is my script

Code:

#!/bin/bash

result=$(ps -ef | grep -v grep | grep "reportgen" | awk '{print $2}')

if [ "$result" == "" ];
then
echo "process is not running"
else
echo "process is running"
fi

Thanks
-Vishnu

valen_tino · 11-01-2010, 08:09 AM

Code:

#!/bin/bash

result=`ps -ef | grep -v grep | grep "applicationname" | awk '{print $2}'`
if [ -z "$result" ];
then
echo "process is not running"
else
echo "process is running"
fi