Bash find and replace string

Question

I am writing a code that will change the IP of VestaCP panel automatically. I need the code to replace the variable $oldIP with a new variable $newIP, so I wrote this piece of code.

sudo perl -pi -e 's/${oldIP}/${newIP}/g' /etc/nginx/conf.d/${oldIP}.conf

But the code doesn't replace the oldIP from oldIP.conf (123.123.123.123.conf).

However, if I try

sudo perl -pi -e 's/123.123.123.123/123.456.123.456/g' /etc/nginx/conf.d/123.123.123.123.conf

It works flawlessly, although I can't use this code since ever servers have a different IP and I can't predict the next IP when it changes.

What am I doing wrong here? Any solutions? Thanks in advance!

P.S. If it helps to know, I am on Ubuntu 16.04.

SLePort · Accepted Answer · 2017-07-29 09:46:34Z

3

Your variables are not expanded between single quotes. Use double quotes instead:

sudo perl -pi -e "s/${oldIP}/${newIP}/g" /etc/nginx/conf.d/${oldIP}.conf

edited Jul 29, 2017 at 9:46

answered Jul 29, 2017 at 9:41

SLePort

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1 Comment

Min Gu Kang Over a year ago

Thanks so much! That solved my problem. I can't believe I made that stupid mistake lol

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