Copy lines from a log file based on timestamp

UNIX for Advanced & Expert Users
1

how to copy lines from a log file based on timestamp.

INFO [11:08:00] (RbrProcessFlifoEventSessionEJB.java:processFlight:274) - E_20080521_110754_967: rbrAciInfoObjects listing complete!
INFO [11:08:00] (RbrPnrProcessEventSessionEJB.java:processFlight:197) - Event Seq: 1647575217; Carrier: UA; Flt#: 0106; Origin: LAX; Dest: O
RD; SchedDepDate: 2008-05-21; Psgr Boarded: true; Pnr Loc Value: ZSQJ3I
INFO [11:08:01] (RbrPnrProcessEventSessionEJB.java:processFlight:197) - Event Seq: 1647575217; Carrier: UA; Flt#: 0106; Origin: LAX; Dest: O
RD; SchedDepDate: 2008-05-21; Psgr Boarded: true; Pnr Loc Value: RXK8XI
INFO [11:08:01] (RbrPnrProcessEventSessionEJB.java:processFlight:197) - Event Seq: 1647575217; Carrier: UA; Flt#: 0106; Origin: LAX; Dest: O
RD; SchedDepDate: 2008-05-21; Psgr Boarded: true; Pnr Loc Value: QGPJBI

Like this i have soo much of data. i need to grep some part of lines based on timestamp [11:08:00 to 12:20:55]

pls help me out..

2

OK, do a 'grep "time stamp here" file-name > filteredLogs.tmp; cp filteredLogs.tmp whateverNameYouLike.log and voila :slight_smile:
The grep for particular pattern will output the lines matching the criteria, and then you redirect the result to a temp file, then you copy this file with a preferred name, HTH.