softwareupdate grep Script Help Using High Sierra I have this set as a cronjob to run once a day. 

I've been struggling to work out how to get grep to find the string in the output. Am fairly new to bash scripting.

Any advice on where I went wrong, and examples are appreciated! 

#!/bin/bash

printf '\e[2t'

check=$(softwareupdate -l)

sleep 5

echo $check


if [ fgrep "No new software" &lt;&lt;&lt; $check ]
then
say "Peter, You are up to date"
else
say "Peter, you have updates"
fi
softwareupdate prints No new software available. on stderr, not on stdout, so it will never get assigned to check. The quick fix is to run

check=$(softwareupdate -l 2&gt;&amp;1)


This will redirect stderr (file descriptor 2) to stdout (file descriptor 1). 

Also your if statement is wrong, the [ is a command, not just syntax. So you can simply run

if fgrep "No new software" &lt;&lt;&lt; $check; then


for this.

But if you don't need the value of $check later on you can put all this in one line and just run

if softwareupdate -l 2&gt;&amp;1 | fgrep -q "No new software"; then