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Top Forums Shell Programming and Scripting Print remaining lines using grep Post 302987271 by rohit_shinez on Wednesday 7th of December 2016 10:50:06 PM
Old 12-07-2016
Print remaining lines using grep

Hi All,

I am having a text file like below

Code:
ERROR - Not a valid
ID : 123


ERROR - Not a valid
hello
ID : 124

SUCCESS - Valid
ID : 12

I need to display like below after reading the file if it finds the error keyword
along with displaying this first line when error pattern is matched

Code:
ERROR - Not a valid
There is Error for ID 123
ERROR - Not a valid
There is Error for ID 124

I have tried something like this but need to print only first line of pattern match and display which ID. There will be always ID printed in text file
Code:
grep -A2 ERROR test.txt

ERROR - Not a valid
ID : 123

--
ERROR - Not a valid
hello
ID : 124

 

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GREP(1) 						      General Commands Manual							   GREP(1)

NAME
grep - search a file for lines containing a given pattern SYNOPSIS
grep [-elnsv] pattern [file] ... OPTIONS
-e -e pattern is the same as pattern -c Print a count of lines matched -i Ignore case -l Print file names, no lines -n Print line numbers -s Status only, no printed output -v Select lines that do not match EXAMPLES
grep mouse file # Find lines in file containing mouse grep [0-9] file # Print lines containing a digit DESCRIPTION
Grep searches one or more files (by default, stdin) and selects out all the lines that match the pattern. All the regular expressions accepted by ed and mined are allowed. In addition, + can be used instead of * to mean 1 or more occurrences, ? can be used to mean 0 or 1 occurrences, and | can be used between two regular expressions to mean either one of them. Parentheses can be used for grouping. If a match is found, exit status 0 is returned. If no match is found, exit status 1 is returned. If an error is detected, exit status 2 is returned. SEE ALSO
cgrep(1), fgrep(1), sed(1), awk(9). GREP(1)
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