Using bash to generate random IPs in Curl

Question

How to use random ip address in Curl request,I'm using this code and worked

printf "%d.%d.%d.%d\n" "$((RANDOM % 256))" "$((RANDOM % 256))" "$((RANDOM % 256))" "$((RANDOM % 256))"

but when use this code in Curl request and test on http://ifconfig.me not worked

curl --header 'X-Forwarded-For: printf "%d.%d.%d.%d\n" "$((RANDOM % 256))" "$((RANDOM % 256))" "$((RANDOM % 256))" "$((RANDOM % 256))"' http://ifconfig.me

Answer 1

Is suggest to use command substitution. Replace

curl --header 'X-Forwarded-For: printf "%d.%d.%d.%d\n" "$((RANDOM % 256))" "$((RANDOM % 256))" "$((RANDOM % 256))" "$((RANDOM % 256))"' http://ifconfig.me

with

curl --header "X-Forwarded-For: $(printf "%d.%d.%d.%d\n" "$((RANDOM % 256))" "$((RANDOM % 256))" "$((RANDOM % 256))" "$((RANDOM % 256))")" http://ifconfig.me
              ^                 ^^                                                                                                      ^^

I switched from '...' to "..." and from printf "..." to $(printf "...").

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See: Difference between single and double quotes in bash