Bash curl HTTP status code as variable

Question

I have a curl call:

curl --silent --connect-timeout 8 --output /dev/null http://0.0.0.0:5000/twi?c0=24 -I -w "%{http_code}\n" ``

It works fine and returns 200 if succeeded. I want the status code in a variable and tried:

curl --silent --connect-timeout 5 --output /dev/null http://0.0.0.0:5000/twi?c0=24

And then:

curl_status=$?
echo $curl_status

This call works well but gives 0 as $curl_status.

I tried:

curl_status=$(curl --silent --connect-timeout 5 --output /dev/null http://0.0.0.0:5000/twi?c0=24)
echo $curl_status

With the same result: call is executed, but gives 0 as $url_status.

Why do I not get the http_code of the first call into a variable to use in an if statement? I read many posts here with no success for a solution.

Your expectation that curl uses the http status as exit code is just wrong. Read the manual if you are unsure if, it is meant for that: man curl. — hek2mgl
– hek2mgl, Commented Jun 28, 2018 at 12:17

lojza · Accepted Answer · 2018-06-28 12:36:12Z

2

Try

curl_status=`curl --silent --connect-timeout 8 --output /dev/null http://0.0.0.0:5000/twi?c0=24 -I -w "%{http_code}\n"`

If you want to follow redirects use -L

answered Jun 28, 2018 at 12:36

lojza

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Comments

heimi · Accepted Answer · 2018-06-28 12:50:30Z

1

With help of the link from Biffen and the example from https://coderwall.com/p/taqiyg/use-http-status-codes-from-curl I came to

curl_status=$(curl -s -o /dev/null -w '%{http_code}' --connect-timeout 3 http://0.0.0.0:5000/twi?c0=24+91+24+91+24+91+24+91)``

Thanks for the help.

answered Jun 28, 2018 at 12:50

heimi

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