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Top Forums Shell Programming and Scripting Get all lines in a file after particular time Post 302348017 by deepakgang on Thursday 27th of August 2009 07:18:03 AM
Old 08-27-2009
Get all lines in a file after particular time

Hi All,

I am lookig for a way to get all the lines from a log file which has been updated 5 mins prior to the system time.

The log file will be like below:

09:01:00 Started polling
09:01:05 Checking directory test
09:02:00 Error! Cannot access directory test
09:03:00 Polling


I get the system time and calculate 5 mins before the system time. For example if the system time is 09:05, then the script will calculate the time to check as 09:00. Then a sed command will be run to get all the lines below the pattern:

sed '/$time/,$p' $logfile

Which will print all the lines below the pattern. But if the log file does not contain 09:00, no lines will be grepped. Is there a way to solve this issue?

Thanks
D
 

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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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