Shell Script to read output of command

Question

I'm doing a script to monitor some things from websites and one of the things is to monitor the http status and their response time.

In the script I run a command to get the http_status: (http command is provided by: httpie : A Curl-like tool for humans)

http -timeout 10 -follow -h http: //$I/ | grep "HTTP\/1.1" | awk '{print $2}'

This command will return the status itself, ie: 200, 404, 403, etc. or will return two other things:

http: error: Request timed out (10.0s)

or

http: error: ConnectionError ..."

Note: Increasing the timeout does not solve my problem. I need it to be 10 seconds.

How do I put a specific code when returning these two other options? For example in the timeout return 9999 and error 8888.

Answer 1

A simple case statement should do what you want:

case "$(http_status)" in
    "case 1")
        return 199
        ;;
    "case 2")
        return 198
        ;;
    *)
        # All is well; do nothing
        ;;
esac

You can use wildcards, e. g. *"timed out"*) or *"ConnectionError"*).