Why does grep not use my variable in this shell script? [duplicate]

Question

#!/bin/sh
#
# found if this IP is receiving data
#
#echo hexIpAdd=`sh convertIPAddressToHex $1`

hexIpAdd='C0 A8 01 02'

#`snmpwalk -v 2c -c public 192.168.252.187 .1.3.6.1.4.1.231.7.1.2.2.1.4.7.13.1.1.3301 | grep '$hexIpAdd' | wc -l`

if [ `snmpwalk -v 2c -c public 192.168.252.187 .1.3.6.1.4.1.231.7.1.2.2.1.4.7.13.1.1.3301 | grep '$hexIpAdd' | wc -l` -gt 0 ]; then
       echo "Value FOUND!!"
else
       echo "Value NOT FOUND!!"
fi

The code is very simple, I convert my an regular IP to a hexadecimal one, and query via SNMPWALK a network equipment to see if the converted IP is receiving streaming.

The problem is, that the grep pipe is failing to use the VARIABLE, with means if I replace the "grep '$hexIpAdd'" with this "grep 'C0 A8 01 02'" it works.

Answer 1

1. Don't use single quote for quoting variable in shell
2. Use grep -q to avoid command substitution, piping with wc and [...] expression:

Have it this way:

hexIpAdd='C0 A8 01 02'

if snmpwalk -v 2c -c public 192.168.252.187 .1.3.6.1.4.1.231.7.1.2.2.1.4.7.13.1.1.3301 | 
    grep -q "$hexIpAdd"; then
       echo "Value FOUND!!"
else
       echo "Value NOT FOUND!!"
fi