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

Can you clarify "not worked" ? It returns the IP – dash-o Aug 30 '20 at 08:48 — dash-o, Aug 30 '20 at 08:48
@dash-o return and show my local Ip – shadi ahmadian Aug 30 '20 at 08:54 — shadi ahmadian, Aug 30 '20 at 08:54

Cyrus · Accepted Answer · 2020-08-30T09:05:42.437

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 "...").

See: Difference between single and double quotes in bash

Using bash to generate random IPs in Curl

1 Answers1