问题
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
来源：https://stackoverflow.com/questions/38642827/bash-how-to-pass-a-line-feed-to-perl-from-within-a-bash-script