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Usually grep command is used to display the line contaning the specified pattern. Is there any way to display n lines before and after the line which contains the specified pattern?
Can this will be achieved using awk?
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To find a pattern that is more than one word long, enclose the string with single or double quotation marks. The grep command can search for a string in groups of files. When it finds a pattern that matches in more than one file, it prints the name of the file, followed by a colon, then the line matching the pattern.
grep understands three different versions of regular expression syntax: ``basic'' (BRE), ``extended'' (ERE) and ``perl'' (PRCE). In GNU grep, there is no difference in available functionality between basic and extended syntaxes.
The grep filter searches a file for a particular pattern of characters, and displays all lines that contain that pattern. The pattern that is searched in the file is referred to as the regular expression (grep stands for global search for regular expression and print out).
Yes, use
grep -B num1 -A num2
to include num1 lines of context before the match, and num2 lines of context after the match.
EDIT:
Seems the OP is using AIX. This has a different set of options which doesn't include -B and -A
this link describes grep on AIX 4.3 (it doesn't look promising)
Matt's perl script might be a better solution.
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