Exit bash script when curl gets a non 200 HTTP status

bashcurlscriptshellshell-script

I have a bash script setup to perform a few curl requests

for currency in EUR INR JPY
do
  curl -i --data '{"currency": "'$currency'"}' -H "Accept: application/json" -H "Content-Type: application/json" http://0.0.0.0:8080/price && echo
done

Is there a way to make the script exit if one of the curl responses come back with an http status != 200?

I also want to keep the standard curl output, e.g I don't want a solution that only prints the http status code.

Cheers

Best Answer

--fail, as mentioned in the man page, seems to do the job:

$ curl --fail --location http://google.com/nope
$ echo $?
22

Related Solutions

Getting curl to output HTTP status code

This should work for you if the web server is able to respond to HEAD requests (this will not perform a GET):

curl -I http://www.example.org

As an addition, to let cURL follow redirects (3xx statuses) add -L.

What’s the exit code for “curl -I” when not HTTP 200

curl -I will always return 0, if it managed to produce an output with the HEAD. You have two alternatives.

The first is to use curl -I --fail instead, and check for exit code 22.

If you're doing this in a Python script, it could look like:

try:
    subprocess.check_call(['curl', '-I', '--fail', url])
except subprocess.CalledProcessError as e:
    if e.returncode == 22:
        (do something)

The second is to actually ask just the HTTP status code, like this:

$ curl -s -I -o /dev/null -w '%{http_code}' $bad-url
403

Best Answer

Related Solutions

Getting curl to output HTTP status code

What’s the exit code for “curl -I” when not HTTP 200

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