如何grep某个模式上下的行

How to grep for lines above and below a certain pattern

我想搜索一个特定的图案(比如条线)，还要在图案上方和下方(即1行)打印线条，或者在图案上方和下方打印2条线条。

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Foo line
Bar line
Baz line

....

Foo1 line
Bar line
Baz1 line

....

使用带参数-A和-B的grep表示在您希望在模式周围打印之前A行和B的行数a：

1	grep -A1 -B1 yourpattern file

An代表"匹配后"的n行。
Bm代表"匹配前"的m行。

如果两个数字相同，只需使用-C：

1	grep -C1 yourpattern file

测试

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$ cat file
Foo line
Bar line
Baz line
hello
bye
hello
Foo1 line
Bar line
Baz1 line

我们grep：

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$ grep -A1 -B1 Bar file
Foo line
Bar line
Baz line
--
Foo1 line
Bar line
Baz1 line

要删除组分隔符，可以使用--no-group-separator：

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$ grep --no-group-separator -A1 -B1 Bar file
Foo line
Bar line
Baz line
Foo1 line
Bar line
Baz1 line

来自man grep：

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-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.