grep for date forward in a syslog

[KPS]

Hi,

This is a grep question. I'm trying to grep for a starting point in a syslog by a specific date and then capture everything else beyond that poing going forward to be able to more it in an active shell and display it. What would the syntax be to do this? I'm doing the following and don't seem to be having any luck with this:

grep "Jul 3" syslog.log | more

Thanks,
KPS

[Ermin Borovac · Accepted solution]

Don't know about grep but you can do it with sed.

$ sed -n '/Jul 3/,$p' syslog.log

This should print everything from the line matching "Jul 3" to the last line ($).

[Vibhor Kumar Agarwal]

If you need you can also pass the search pattern as a variable


sed -n "/$Search_Pattern/,$p" syslog.log

Just remember to use double quotes (") instead of single one (')

[Muthukumar_5]

#!/usr/bin/ksh
LINE=$(awk '/Jul 30/{ print NR;exit; }' syslog.log)
tail -n +${LINE} syslog.log

hth

[john korterman]

Hi,
perhaps you can use the awk funtionality for printing everything between a starting and an end point, e.g.:

# awk '/Jul 6/,/\$/' /var/adm/syslog.log|more

(there are two spaces beteen "Jul" and "6" in the above example).

regards,
John K.