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

问题:

This question already has an answer here:

• How to replace multiple any-character (including newline) in Perl RegEx? 1 answer

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?

回答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