如何grep在特定模式上方和下方的线条 [英] How to grep for lines above and below a certain pattern
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问题描述
我想搜索某个图案(比如条形线),但还要在图案上方和下方(即 1 行)或图案上方和下方打印 2 行.
I would like to search for a certain pattern (say Bar line) but also print lines above and below (i.e 1 line) the pattern or 2 lines above and below the pattern.
Foo line
Bar line
Baz line
....
Foo1 line
Bar line
Baz1 line
....
推荐答案
使用 grep
和参数 -A
和 -B
来表示A
after 和 B
之前要在图案周围打印的行数 a:
Use grep
with the parameters -A
and -B
to indicate the number a of lines A
fter and B
efore you want to print around your pattern:
grep -A1 -B1 yourpattern file
An
代表匹配之后"的n
行.Bm
代表匹配之前"的m
行.An
stands forn
lines "after" the match.Bm
stands form
lines "before" the match.
如果两个数字相同,就用-C
:
If both numbers are the same, just use -C
:
grep -C1 yourpattern file
测试
$ cat file
Foo line
Bar line
Baz line
hello
bye
hello
Foo1 line
Bar line
Baz1 line
让我们grep
:
$ grep -A1 -B1 Bar file
Foo line
Bar line
Baz line
--
Foo1 line
Bar line
Baz1 line
要去掉组分隔符,可以使用--no-group-separator
:
To get rid of the group separator, you can use --no-group-separator
:
$ grep --no-group-separator -A1 -B1 Bar file
Foo line
Bar line
Baz line
Foo1 line
Bar line
Baz1 line
来自man grep
:
-A NUM, --after-context=NUM
Print NUM lines of trailing context after matching lines.
Places a line containing a group separator (--) between
contiguous groups of matches. With the -o or --only-matching
option, this has no effect and a warning is given.
-B NUM, --before-context=NUM
Print NUM lines of leading context before matching lines.
Places a line containing a group separator (--) between
contiguous groups of matches. With the -o or --only-matching
option, this has no effect and a warning is given.
-C NUM, -NUM, --context=NUM
Print NUM lines of output context. Places a line containing a
group separator (--) between contiguous groups of matches. With
the -o or --only-matching option, this has no effect and a
warning is given.
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