sed怎么 删除匹配区间的所有行!

筷子eason

[root@1 shell]# cat sed.txt
eason chan
520
eason
131
chen
520
789
334

要求：必须用sed来 删除文本520开头到131结尾的所有行
结果如下：
eason chan
chen
520
789
334

moperyblue

1. sed '0,/^520/{//d;b};0,/131$/d'
   
2. sed '/^520/{:a;N;/131$/!ba;d}'

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筷子eason

回复 2# moperyblue

非常感谢  写的非常的好
就是第一个看的不是很懂，能否简单的解析一下呢

moperyblue

回复 3# 筷子eason


info sed
=>
`0,/REGEXP/'
     A line number of `0' can be used in an address specification like
     `0,/REGEXP/' so that `sed' will try to match REGEXP in the first
     input line too.  In other words, `0,/REGEXP/' is similar to
     `1,/REGEXP/', except that if ADDR2 matches the very first line of
     input the `0,/REGEXP/' form will consider it to end the range,
     whereas the `1,/REGEXP/' form will match the beginning of its
     range and hence make the range span up to the _second_ occurrence
     of the regular expression.

     Note that this is the only place where the `0' address makes
     sense; there is no 0-th line and commands which are given the `0'
     address in any other way will give an error.

`/REGEXP/'
     This will select any line which matches the regular expression
     REGEXP.  If REGEXP itself includes any `/' characters, each must
     be escaped by a backslash (`\').

     The empty regular expression `//' repeats the last regular
     expression match (the same holds if the empty regular expression is
     passed to the `s' command).  Note that modifiers to regular
     expressions are evaluated when the regular expression is compiled,
     thus it is invalid to specify them together with the empty regular
     expression.

sunzhiguolu

1. perl -0pe 's/^520(?s:.*?131\n)//m' f

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筷子eason

回复 5# sunzhiguolu

牛

筷子eason

回复 4# moperyblue

谢谢啊  info的确详细

moperyblue

1. awk '/^520/&&!p{p=1;f=1}f{if(/131$/)f=0;next}1'

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