How to use Curl to also output first 20 characters from the response? - bash

Question

I'm using curl to retrieve the http_code size_header redirect_url and Website Title with:
#!/bin/bash
FILE="$1"
while read LINE; do
curl -H 'Cache-Control: no-cache' -i -s -k -o >(perl -l -0777 -ne 'print $1 if /<title.*?>\s*(.*?)\s*<\/title/si') --silent --max-time 2 --write-out '%{http_code} %{size_header} %{redirect_url} ' "$LINE"
echo " $LINE"
done < ${FILE}
but I like to also retrieve the first 20 characters from the response to have more information.
The idea is to get this output
%{http_code} %{size_header} %{redirect_url} $website_title $website_first_20_bytes
I only need to add the $website_first_20_bytes to the output. How can I achieve this?
PS: No the first 20 characters from header response. Only the source.

Answer 1

So you probably mean something like this then (I added a bunch of newlines etc. to the output which you can trim as you please):
#!/bin/bash
FILE="$1"
while read -r LINE; do
# read the response to a variable
response=$(curl -H 'Cache-Control: no-cache' -s -k --max-time 2 --write-out '%{http_code} %{size_header} %{redirect_url} ' "$LINE")
# get the title
title=$(sed -n 's/.*<title>\(.*\)<\/title>.*/\1/ip;T;q'<<<"$response")
# read the write-out from the last line
read -r http_code size_header redirect_url < <(tail -n 1 <<<"$response")
printf "Status: %s\n" "$http_code"
printf "Size: %s\n" "$size_header"
printf "Redirect-url: %s\n" "$redirect_url"
printf "Url: %s\n" "$LINE"
printf "Title: %s" "$title"
# -c 20 only shows the 20 first chars from response
printf "Body: %s" "$(head -c 20 <<<"$response")"
done < ${FILE}