打印两个模式之间的所有行，仅排他，仅第一实例(在sed，AWK或Perl中) [英] Print all lines between two patterns, exclusive, first instance only (in sed, AWK or Perl)

问题描述

使用sed，AWK(或Perl)，如何打印两个模式(第一个实例)之间的所有行(不包括模式)? ¹

Using sed, AWK (or Perl), how do you print all lines between (the first instance of) two patterns, exclusive of the patterns?¹

也就是说，作为输入给出:

That is, given as input:

aaa
PATTERN1
bbb
ccc
ddd
PATTERN2
eee

甚至可能:

aaa
PATTERN1
bbb
ccc
ddd
PATTERN2
eee
fff
PATTERN1
ggg
hhh
iii
PATTERN2
jjj

在两种情况下，我都会期望:

I would expect, in both cases:

bbb
ccc
ddd

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¹ _{许多用户投票结束了该问题，并重复了要点，以证明它们是不同的.这个问题从表面上也类似于 a}

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¹ _{A number of users voted to close this question as a duplicate of this one. In the end, I provided a gist that proves they are different. The question is also superficially similar to a number of others, but there is no exact match, and none of them are of high quality, and, as I believe that this specific problem is the one most commonly faced, it deserves a clear formulation, and a set of correct, clear answers.}

推荐答案

使用 awk (假设 PATTERN1 和 PATTERN2 始终存在配对，而它们中的任何一个都不会出现在配对中)

With awk (assumes that PATTERN1 and PATTERN2 are always present in pairs and either of them do not occur inside a pair)

$ cat ip.txt
aaa
PATTERN1
bbb
ccc
ddd
PATTERN2
eee
fff
PATTERN1
ggg
hhh
iii
PATTERN2
jjj

$ awk '/PATTERN2/{exit} f; /PATTERN1/{f=1}' ip.txt
bbb
ccc
ddd

如果/PATTERN1/与之匹配，则
• /PATTERN1/{f = 1} 设置标志
如果/PATTERN2/匹配，则
• /PATTERN2/{exit} 退出
• f; 如果设置了标志，则打印输入行
• /PATTERN1/{f=1} set flag if /PATTERN1/ is matched
• /PATTERN2/{exit} exit if /PATTERN2/ is matched
• f; print input line if flag is set


通用解决方案，可以在其中指定所需的块


Generic solution, where the block required can be specified

$ awk -v b=1 '/PATTERN2/ && c==b{exit} c==b; /PATTERN1/{c++}' ip.txt
bbb
ccc
ddd
$ awk -v b=2 '/PATTERN2/ && c==b{exit} c==b; /PATTERN1/{c++}' ip.txt
2
46

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