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
– dash-o

2020-08-30 08:48:22 +00:00
Commented Aug 30, 2020 at 8:48 — dash-o
– dash-o, Commented Aug 30, 2020 at 8:48
@dash-o return and show my local Ip

shadi ahmadian
– shadi ahmadian

2020-08-30 08:54:01 +00:00
Commented Aug 30, 2020 at 8:54 — shadi ahmadian
– shadi ahmadian, Commented Aug 30, 2020 at 8:54

Cyrus · Accepted Answer · 2020-08-30 09:05:42Z

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 "...").

See: Difference between single and double quotes in bash

edited Aug 30, 2020 at 9:05

answered Aug 30, 2020 at 8:49

Cyrus

90.2k15 gold badges112 silver badges173 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

Using bash to generate random IPs in Curl

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related