Printing HTTP status code with requestsed url

Question

I am using a tool called waybackurls which is returning a list of URLs.

waybackurls api.target.com | grep "url"

Output

https://api.target.com/ws/api/upg/customer-bff/v2/rest/url?=value

Now I am trying to validate the return URL through curl. Because it returns lot of false-positive results. I am able get the http status code only

waybackurls api.target.com | grep "url" | xargs -n 1 curl -s -o /dev/null -w "%{http_code}"

Output

what I want output is using curl

200  > https://api.target.com/ws/api/upg/customer-bff/v2/rest/url?=value
404  > https://api.target.com/ws/api/redirect_to?=value

Is it possible to get this formate with curl request?

thanks

Saboteur · Accepted Answer · 2020-04-16 08:59:58Z

1

Just check curl manual. You already put http_code, the same for url:

waybackurls api.target.com | grep "url" | xargs -n 1 curl -s -o /dev/null -w "%{http_code} > %{url_effective}\n"

answered Apr 16, 2020 at 8:59

Saboteur

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