Perl one-liner regex doesn't substitute correctly - returns error

This is probably something kick-self obvious but I have to ask as I'm not seeing it.

I'm trying to make the following substitution in this text file. I want this:

bind_password = 'grafana'

to become this:

bind_password = ''

I've tried using the following regex one-liner on the file that contains the line:

$ perl -0pe 's/(bind_password = \')grafana\'/$1\'/g' file.txt
bash: syntax error near unexpected token `)'

When I've tried the regex on regex101, it's worked fine: https://regex101.com/r/0fb4Pu/1. The difference is that I've had to escape the single quotes. I've tried using double quotes instead of single around the regex, as in:

perl -0pe "s/(bind_password = ')grafana'/$1'/g" file.txt

But while this doesn't return an error, it doesn't do what I want. It instead replaces the string with a single colon, like so:

What am I doing wrong here?

Solution

You can use

perl -i -pe "s/bind_password = '\Kgrafana(?=')//" file.txt

Details:

-i - modify the file contents
bind_password = '\K - match bind_password = ' and remove this text from the match value (so that it stays intact in the file)
grafana - your grafana word
(?=') - followed with ' (but ' is not consumed, so it stays in the file intact)

Note:

You do not need 0 in -0pe because your match is not spanning multiple lines
Escaping single quotes does not work like this, it is simpler to use dobule quotes here around the expression, although sometimes concatenation is the only way to go
You need no g flag since this perl works on a line-by-line basis, and processes all lines (g is necessary when there are multiple matches per line).