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Full Discussion: awk to print specific line in file based on criteria
Top Forums Shell Programming and Scripting awk to print specific line in file based on criteria Post 302982047 by Don Cragun on Thursday 22nd of September 2016 03:31:54 PM
Old 09-22-2016
Hi cmccabe,
If I correctly understand your specification, you could also try this alternative using your original input field separator specification:
Code:
awk -F"\"[,:]* *\"*" '
{	for(i = 2; i < NF - 3; i++)
		if($i       == "path" &&
		   $(i + 2) == "plugin" &&
		   $(i + 3) == "/rundb/api/v1/plugin/49/")
			print $(i + 1)
}
' file

(and I must say that that is a nicely constructed field separator ERE). You could simplify it a little bit using single quotes to avoid needing to escape the double quotes: -F'"[,:]* *"*'.

Hi R. Singh,
Note that using -F"[\"|:|,]" specifies that any ", |, :, |, or , character is to be treated as a field separator because every character in a bracket expression (other than the backslash used to escape a double-quote in a shell double-quoted string) is a match. There is no alternation in a bracket expression. You could get the same effect (unless you really wanted | to be a field separator) with -F'[":,]' (using a matching bracket expression) or with -F'"|:|,' (using alternation without a bracket expression). And, if there were any | characters in the input, your expression would interpret them as field separators instead of as data.
Last edited by Don Cragun; 09-23-2016 at 05:13 AM.. Reason: Fix typo: s/needed/needing/
These 2 Users Gave Thanks to Don Cragun For This Post:
cmccabe RavinderSingh13
 

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Plugin(3pm)						  LogReport's Lire Documentation					       Plugin(3pm)

NAME

       Lire::Config::Plugin - Value object for plugin configuration.

SYNOPSIS

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