如何使用bash脚本在文件中输出第一行curl结果

Question

I want to read a file of urls, curl each url and only get the first line which contains the HTTP code. I am running under Windows 10 inside Cmder.

#!/bin/bash
input="urls.csv"
truncate -s 0 dest.csv
while IFS= read -r var
do
    result= `curl -I ${var%$'\r'} | grep HTTP $result`
echo "$var $result" >> dest.csv
done < "$input"

however the output file is empty, thank you

Answer 1

Assuming urls.csv is just a simple list of URLs and you're working on a linux system (or any system which has /dev/null ), following command will send HEAD requests to each URL and output them next to HTTP response codes.

sed 's/^/url = /; s/\r\?$/\n-o \/dev\/null/' urls.csv |
curl -s -K- -w '%{http_code} %{url_effective}\n' -I >outfile

see curl man page for further information.

Answer 2

In reply to your query on outputting the HTTP code, this is given in the second line of the header. You may get this by:

export IFS=$; # This is required to stop everything coming out on one line!
curl -i <domain> | head -n 2