使用awk更改与分隔符之间的模式匹配的行

Question

我有一个像这样的输入文件:

some line
some other line
another line
start_delimiter
  interesting stuff
  more interesting stuff
  even more interesting stuff
end_delimiter
possibly more stuff

我想操纵start_delimiter和end_delimiter之间与正则表达式匹配的行,并将结果写入输入文件.例如,将'//'添加到包含单词'more'的行的开头(只要这些行在分隔符之间):

some line
some other line
another line
start_delimiter
  interesting stuff
  //more interesting stuff
  //even more interesting stuff
end_delimiter
possibly more stuff

我可以通过这种方式在分隔符之间获取文本部分:

awk '/start_delimiter/,/end_delimiter/' inputfile

如果我把它管道到另一个awk我可以改变我感兴趣的行:

awk '/more/ {sub(/^/,"//")}1'

我不确定的是如何返回并用新内容替换分隔部分.有任何想法吗？单线的奖励积分.

Answer 1

/start_delimiter/ {inblock=1}
/end_delimiter/ {inblock=0;}
{ if (inblock==1 && (/more/)) { print "//" $0 } else print $0}

Answer 2

而对于一个队员狂热分子:

awk '/^(start_delim|end_delim)/{f=f?0:1}f&&/more/{$0="//" $0}1' file

^如果块分隔符嵌入在行的中间,则添加行锚的开头以避免混淆.