I have tried this command

curl -Ik https://dev.mydomain.com/

and it does print everything. And now what I want is to print out content-security-policy only.

Do I need to use jq or is there any other helpful tool that I can use?

CodePudding user response：

curl  -sIk https://stackoverflow.com/ | grep content-security-policy | cut -d ' ' -f 2-

Will curl the url, grep only the line with content-security-policy, cut on a space, and get all the fields from 2 onwards.

Example:

➜  ~ curl  -sIk https://stackoverflow.com/ | grep content-secur | cut -d ' ' -f 2-
upgrade-insecure-requests; frame-ancestors 'self' https://stackexchange.com

CodePudding user response：

If you use cURL >= 7.84.0, you can use the syntax %header{name} :

curl -Iks https://stackoverflow.com -o /dev/null -w "%header{content-security-policy}"

If you want to try it without installing a new version, you can run the Docker image :

docker run --rm curlimages/curl:7.85.0 -Iks https://stackoverflow.com -o /dev/null -w "%header{content-security-policy}"