Linux 如何让 grep 打印每个匹配行下方和上方的行？
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How can I make grep print the lines below and above each matching line?
提问by poiuytrez
I have to parse a very large file and I want to use the command grep (or any other tool).
我必须解析一个非常大的文件，并且我想使用命令 grep（或任何其他工具）。
I want to search each log line for the word FAILED, then print the line above and below each matching line, as well as the matching line.
我想在每个日志行中搜索单词FAILED，然后在每个匹配行的上方和下方打印该行，以及匹配的行。
For example:
例如：
id : 15
Satus : SUCCESS
Message : no problem
id : 15
Satus : FAILED
Message : connection error
And I need to print:
我需要打印：
id : 15
Satus : FAILED
Message : connection error
采纳答案by pgs
grep's -A 1option will give you one line after; -B 1will give you one line before; and -C 1combines both to give you one line both before and after, -1does the same.
grep 的-A 1选项后会给你一行；-B 1之前会给你一行；并-C 1结合两者在前后给你一行，-1做同样的事情。
回答by Milan Babu?kov
Use -A and -B switches (mean lines-after and lines-before):
使用 -A 和 -B 开关（平均后行和前行）：
grep -A 1 -B 1 FAILED file.txt
回答by tefozi
Use -B, -A or -C option
使用 -B、-A 或 -C 选项
grep --help
...
-B, --before-context=NUM print NUM lines of leading context
-A, --after-context=NUM print NUM lines of trailing context
-C, --context=NUM print NUM lines of output context
-NUM same as --context=NUM
...
