Bash script to test website status, then run command-CodePudding

I am trying to write a very basic bash script, to run on a linux server. I am not sure why it's not working.

The intention is to test if a website returns a 200 OK response. If it does, then exit. If it does not, then run a command.

The script is:

#!/bin/bash

if HEAD https://google.com | grep '200\ OK' | wc -l;  then
   echo "Site is up";
else
   sudo wo clean --all && sudo wo stack reload --all
fi

The issue is that if the site is up, it gives the expect output. Namely ...

1
Site is up

But if the site is down (say I replace https://google.com with https://766google.com, the output I get is:

0
Site is up

I've tried numerous variations, including put the HEAD ... command in [ ].

For example, I tried this:

#!/bin/bash

if [ HEAD https://google.com | grep '200\ OK' | wc -l ];  then
   echo "Site is up";
else
   sudo wo clean --all && sudo wo stack reload --all
fi

But then when the site is down, it runs the desired command, but gives an error ... I get this output:

./sitecheck.sh: line 3: [: missing `]'
wc: ]: No such file or directory
Cleaning NGINX FastCGI cache ... 
[THE OUTPUT OF COMMAND IT RUNS WHEN SITE IS DOWN]

Would someone please explain what is wrong with this script?

Thanks.

CodePudding user response：

If you want to count the number of lines of output, you need to capture the pipeline's output with $(...), and then use ((...)) to do an arithmetic comparison. One set of parentheses is output capturing; two sets is arithmetic.

if (($(HEAD https://google.com | grep '200 OK' | wc -l) > 0)); then
   echo "Site is up";
else
   sudo wo clean --all && sudo wo stack reload --all
fi

A simpler way to do this is to use grep -q. Instead of outputting matches, requiring you to parse grep's output, grep -q directly returns a success or failure exit code indicating whether it found a match.

if HEAD https://google.com | grep -q '200 OK'; then
   echo "Site is up";
else
   sudo wo clean --all && sudo wo stack reload --all
fi