如何在19:55的grep文件中获得第1,2,3,4,5行？
2013/10/08 19:55:27.471 Line 1 Line 2 Line 3 Line 4 Line 5 2013/10/08 19:55:29.566 Line 1 Line 2 Line 3 Line 4 Line 5
你要：
原文链接：https://www.f2er.com/bash/388248.htmlgrep -A 5 '19:55' file
从人grep：
Context Line Control -A NUM,--after-context=NUM Print NUM lines of trailing context after matching lines. Places a line containing a gup separator (described under --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 (described under --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 (described under --group-separator) between contiguous groups of matches. With the -o or --only-matching option,this has no effect and a warning is given. --group-separator=SEP Use SEP as a group separator. By default SEP is double hyphen (--). --no-group-separator Use empty string as a group separator.
