grep return 0 if no match

Question

Is there any bash script which can return me a result=true with command grep?

Example: There are 1000 records of 103.12.88 in my CF logs. Can I do a grep 103.12.88 if detect 1 or more results then print/output result show me either YES or True

Answer 1

The actual answer to this question is to add || true to the end of the command, e.g.:

echo thing | grep x || true

This will still output a 0 return code.

Answer 2

You are processing the return value incorrectly.

value=$( grep -ic "210.64.203" /var/logs )

sets value to the output of the grep, not to its return code.

After executing a command the exit code it stored in $?, but you usually don't need it.

if grep -ic "210.64.203" /var/logs 
then echo "Found..."
else echo "not found"
fi

If you want the value, then test for content.

rec="$( grep -ic "210.64.203" /var/logs )"
if [ -n "$rec" ] ; then echo found; fi

Or if using bash,

if [[ "$rec" ]] ; then echo found; fi

though I prefer to be explicit -

if [[ -n "$rec" ]] ; then echo found; fi

Answer 3

[user@host ~]$ cat ~/mylogfile.txt
no secrets here
[user@host ~]$ grepnot(){ ! grep $1 $2; return $?;}
[user@host ~]$ grepnot password mylogfile.txt 
[user@host ~]$ echo $?
0
[user@host ~]$ grepnot secret mylogfile.txt 
no secrets here
[user@host ~]$ echo $?
1

Note: I'm using 'grepnot' in a Jenkins pipeline. My previous answer used "1 - $?" to reverse the return code. But that solution still caused grep failure in the pipeline. The above solution (which borrows from a previous "!" answer) works with the pipeline.