Sed is taking the file name instead of empty value

Question

I have a replace function like

function replace()
{
sed -i "s#$1#$2#g" $3
}

I am calling the function with these parameters

replace MY_IP $MY_IP /usr/xxx.sh

where $MY_IP is a empty value so sed is giving as sed -i s#MY_IP#/usr/xxx.sh#g

sed no input files

It is not taking the empty value. How to solve this?

http://mywiki.wooledge.org/BashPitfalls#cp_.24file_.24target — melpomene, Oct 03 '18 at 11:22

score 0 · Answer 1 · answered Oct 03 '18 at 11:26

0

Try quoting $MY_IP.

replace MY_IP "$MY_IP" /usr/xxx.sh

Without the quotes, bash will just skip that argument.

answered Oct 03 '18 at 11:26

Jon

Thank u so much :) – Heisenberg Oct 03 '18 at 12:59
1

@Heisenberg Additionally, `$3` in `replace` should be quoted. – Benjamin W. Oct 03 '18 at 13:26

1 Answers1