用 sed 和 perl 中的多行文本替换多行文本

Question

Here is my file

      start
      exit 0

      status
      exit 0

      stop
      exit 0

the result file should be

      start
      exit 0

      status
      exit 152

      stop
      exit 0

any help is appreciated How can I do it using sed and perl

Answer 1

Using Awk

Input

$ cat f
      start
      exit 0

      status
      exit 0

      stop
      exit 0

Output

$ awk '/exit/ && p {sub($NF,"152")}{p=/status/}1'  f
      start
      exit 0

      status
      exit 152

      stop
      exit 0

Explanation

• p=/status/ Set variable p true whenever awk finds word status
• /exit/ && p when awk finds word exit on current record and variable p evaluates true then

sub(regexp, replacement [, target])

Search target, which is treated as a string, for the leftmost, longest substring matched by the regular expression regexp. Modify the entire string by replacing the matched text with replacement. The modified string becomes the new value of target. Return the number of substitutions made (zero or one). If 3rd argument is omitted, then the default is to use and alter $0 .

• sub($NF,"152")

Substitute last field ( $NF ) with 152 of current record, since 3rd argument is not given hence, default $0 is altered.

• } 1

1 always evaluates to true, it performs default operation {print $0}

Answer 2

perl -pe's/exit \K\d+/152/ if $f; $f = /status/'

Usage:

perl -i~ -pe'...' file            # Edit in place with backup
perl -i -pe'...' file             # Edit in place without backup
perl -pe'...' file.in >file.out   # Read from file
perl -pe'...' <file.in >file.out  # Read from stdin

Answer 3

With sed one can use:

$ sed '/status/{n;s/0/152/}' input
      start
      exit 0

      status
      exit 152

      stop
      exit 0