Grep status code from curl output

Question

My curl output in console is like below. There will be many curl commands in the shell script which i will be looping through "n" times. I would like to capture/grep only those status which are not equal to 200 OK and pass it to a file. Please suggest me.

HTTP/1.1 200 OK
Authorization: Bearer 2d141ec6-1ac7-458a-96f3-318af10ae3b9

anubhava · Accepted Answer · 2014-12-01 06:25:07Z

2

You can pipe curl's output to this awk:

awk '$2 != 200'

Alternatively you can use this curl command to just get status:

curl -Is -w "%{http_code}" -A "Chrome" -L "http://domain.com" -o /dev/null

To write all the URLs with non 200 in output:

url='http://domain.com/'
[[ $(curl -s -w "%{http_code}" -A "Chrome" -L "$url" -o /dev/null) != 200 ]] && echo "$url"

edited Dec 1, 2014 at 6:25

answered Dec 1, 2014 at 5:44

anubhava

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4 Comments

Jams Over a year ago

my curl is like this- for ((i=1; i<=2; i++)); do curl -i -vs POST -H "$SESSION_TOKEN" -H "Accept:$ACCEPT_HEADER" "http://$BASE_URI/api/$PLATFORM/curated/featured" | awk '$2 != 200' >> RESULT.txt done >> result1.txt

Jams Over a year ago

This is resulting soo many other stuffs in the RESULT.txt file. I just want to have only status code in that file which is not equal to 200

anubhava Over a year ago

In that use my 2nd suggested command that only prints status i.e. 200 or 404.

Freedo Over a year ago

This results code 200 for everything, even links that should be 403

nu11p01n73R · Accepted Answer · 2014-12-01 05:48:18Z

1

The -v option inverts the match

$ grep -v "^HTTP/1.1 200"

answered Dec 1, 2014 at 5:48

nu11p01n73R

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