打印线的具体数量匹配模式之后 [英] print specific number of lines after matching pattern
问题描述
我从输入文件中的前pressionAAA的每次出现后打印81行。我怎么去的?
I have to print 81 lines after each occurrence of the expression "AAA" from my input file. How do I go about that?
推荐答案
下面的成语描述了如何选择给予一定范围的记录
匹配一个特定的模式:
The following idioms describe how to select a range of records given a specific pattern to match:
a)由某种模式打印的所有记录:
a) Print all records from some pattern:
awk '/pattern/{f=1}f' file
二)打印所有记录后,一些模式:
b) Print all records after some pattern:
awk 'f;/pattern/{f=1}' file
C)打印的第N个记录后,一些模式:
c) Print the Nth record after some pattern:
awk 'c&&!--c;/pattern/{c=N}' file
D)打印每条记录后,除了一些模式的第N个记录:
d) Print every record except the Nth record after some pattern:
awk 'c&&!--c{next}/pattern/{c=N}1' file
E)经过一些图案打印N条记录:
e) Print the N records after some pattern:
awk 'c&&c--;/pattern/{c=N}' file
F)打印每条记录后,除了一些模式的N条记录:
f) Print every record except the N records after some pattern:
awk 'c&&c--{next}/pattern/{c=N}1' file
g)因某种模式打印N条记录:
g) Print the N records from some pattern:
awk '/pattern/{c=N}c&&c--' file
我是从F改为变量名的发现,以C为计数,其中
适当因为这是一个什么样的变量实际上是比较前pressive。
I changed the variable name from "f" for "found" to "c" for "count" where appropriate as that's more expressive of what the variable actually IS.
所以,你想要的e上面:
So, you'd want "e" above:
awk 'c&&c--;/AAA/{c=81}' file
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