bash - How to pass a line feed to perl from within a bash script?

Question

I am trying to replace all digits sequences followed by a line feed and the letter

a

. Those digits sequences are located in a file called

test.txt

. The bash script

command.sh

is used to perform the task (see below).

test.txt

00

a1

b2

a

command.sh

#!/bin/bash



MY_VAR="\d+

a"



grep -P "^.+$" test.txt | perl -pe "s/$MY_VAR/D/";

When I call the

command.sh

file, here is what I see:

$ ./command

00

a1

b2

a

However, I'm expecting this output:

$ ./command

D1

bD

What am I missing?

Answer 1

You don't even need grep since it is just matching .+, just use perl with -0777 option (slurp mode) to match across the lines:

#!/bin/bash

MY_VAR="\d+
a"

perl -0777pe "s/$MY_VAR/D/g" test.txt

Output:

D1
bD

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