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LanceBaynes
Asked: 2011-01-25 23:12:52 +0800 CST

Output only the last 2 minute's log

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I have a log file:
http://pastebin.com/raw.php?i=cdUpTqNV

the first column, before the "ASDF" is the same time, as in the given row, but in unix time [i think it's easier to do this with unix time].

I just need the lines what was in the last 2 minutes:
Ending: 07:55:08 - the last time in the logfile
Starting: 07:53:09 - at most 2 minutes before the last log line in the logfile

The output of the pastebined text would be:
1295938389ASDF 01 25 07:53:09 router authpriv.notice dropbear[20673]: password auth succeeded for 'root' from 192.168.1.201:43822
1295938401ASDF 01 25 07:53:21 router daemon.info dnsmasq-dhcp[1140]: DHCPREQUEST(br-lan) 192.168.1.201 01:2c:23:c3:32:f3
1295938401ASDF 01 25 07:53:21 router daemon.info dnsmasq-dhcp[1140]: DHCPACK(br-lan) 192.168.1.201 01:2c:23:c3:32:f3
1295938508ASDF 01 25 07:55:08 router daemon.info dnsmasq-dhcp[1140]: DHCPREQUEST(br-lan) 192.168.1.201 01:2c:23:c3:32:f3
1295938508ASDF 01 25 07:55:08 router daemon.info dnsmasq-dhcp[1140]: DHCPACK(br-lan) 192.168.1.201 01:2c:23:c3:32:f3

How to do this using only minimal *nix tools? [it's an OpenWrt router, no perl :( ]

so, how to do this: output only line(Last log line time - 120sec)?

Thank you!

linux openwrt

2 Answers

  1. Best Answer

    Give this a try:

    last=$(sed -n '$s/^\([0-9]\+\).*/\1/p')
awk -F ASDF -v last=$last '$1 >= last - 120'
    • 1

  2. You can get current timestamp like this:

    date +s%

    And current timestamp - 2mins for example this way:

    date -d @$((`date +%s`-120)) +%s

    So you have two integer values and you want to grep lines which starts within this range. There are many ways how to do it (awk, sed,...)

    Sed example:

    sed -n -e '/start_time_stamp/,/stop_time_stamp/p' log_file

    For other ways, try this magnificent tool: http://lmgtfy.com/?q=grep+number+range

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