我需要使用 sed 查找匹配项并删除该匹配项之前的 2 行和之后的 3 行

Question

我需要使用“sed”找到匹配项，并在该匹配项之前删除 2 行，之后删除 3 行，然后打印输出，我该怎么做？

Answer 1

if the file is not huge, try this:

    awk 'NR==FNR{if($0~/matchWord/){for(i=NR-2;i<=NR+3;i++){if(i!=NR)a[i]++}}}\
NR>FNR{if(!(FNR in a))print $0}' file file

I didn't test, but should work.

Answer 2

First off, you do not want to do this in sed. 2nd, your question is ill posed: what do you do if you have a match on lines 5 and 8? Does line 8 get deleted and line 6 is kept? Assuming that's not a concern, this seems to do what you want:

#!/bin/sed -nf

1{ h; d; } 
H
2,5d
g
/^\([^\n]*\n\)\{2\}match/!P
/^\([^\n]*\n\)\{2\}match/{
  s/\n[^\n]*$//
  N
}
s/[^\n]*\n//
h
$p

Note: if the match occurs in the last 3 lines of the file, this does not behave as desired. That case is left as an exercise for the (masochistic) reader.

Answer 3

sed ‘/matchWord/,+3d;:flag;1,2!{P;N;D};N;bflag’ file