How can I make grep print the lines below and above each matching line?
Possible Duplicate:
grep a file, but show several surrounding lines?
I have to parse a very large file and I want to use the command grep (or any other tool).
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.
For example:
id : 15
Satus : SUCCESS
Message : no problem
id : 15
Satus : FAILED
Message : connection error
And I need to print:
id : 15
Satus : FAILED
Message : connection error
grep's -A 1 option will give you one line after; -B 1 will give you one line before; and -C 1 combines both to give you one line both before and after.
使用-B和-A选项
grep --help
...
-B, --before-context=NUM print NUM lines of leading context
-A, --after-context=NUM print NUM lines of trailing context
...
使用-A和-B开关（之前和之后的平均线）：
grep -A 1 -B 1 FAILED file.txt
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