grep 查找字符串，获取前后行，不显示行号

Question

最近我需要在文件中搜索特定字符串，并且需要显示匹配项之前的行和之后的行。

文件

AAAABBBB                      CCCCDDDD

Using google I found the following solution utilizing grep.

grep -n1 BBBB File

哪些输出，

1-AAAA
2-BBBB
3-CCCC

所以这符合我想要的，除了它在输出中显示行号。

有没有办法抑制行号？

Answer 1

使用这个:

grep -C 1 BBBB input

grep 具有三个用于显示上下文行的选项:

  -A NUM, --after-context=NUM
          Print NUM lines of trailing context after matching lines.  Places a line  containing  --  between  contiguous  groups  of
          matches.
  -B NUM, --before-context=NUM
          Print  NUM  lines  of  leading  context  before matching lines.  Places a line containing -- between contiguous groups of
          matches.
  -C NUM, --context=NUM
          Print NUM lines of output context.  Places a line containing -- between contiguous groups of matches.