Unix Command to grep a time range

Question

I have a log file, saved with particular date. I wanna fetch log entries during a particular time and date range to another file.

Ex: all entries from 2014-12-04 00:00:00 time to 2014-12-04 17:00:00

Solution

Try egrep:

pttrn="2014-12-04 0[0-9]"
pttrn="${pttrn}|2014-12-04 1[0-6]"
pttrn="${pttrn}|2014-12-04 17:00:00"

egrep "${pttrn}" <logfile>

The egrep pattern contains three parts. The first part grabs everything from 00:00:00 to 09:59:59. The second part grabs everything from 10:00:00 to 16:59:59, and the third part grabs 17:00:00.

Comments

[wurtel]

True; you could leave out the :00 part, but then again there would have to be at least one log line per hour. Personally I'd adapt the script to fit the data; perhaps even apply some perl foo.

[G-Man Says 'Reinstate Monica']

Your first egrep eliminates values beginning with 08 or 09 (i.e., times between 08:00 and 09:59), and the second one allows 17:10, 17:20, 17:30, etc..., to get through.

[Daniel Goldfarb]

@G-Man Thanks. You're right. I think I've fixed it now.

[G-Man Says 'Reinstate Monica']

That looks like it should work. It could be simplified to egrep "2014-12-04 (0[0-9]|1[0-6]|17:00:00)".

[Community]

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