linux - Output only the last 2 minute's log

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I have a log file:
href="http://pastebin.com/raw.php?i=cdUpTqNV" rel="nofollow
noreferrer">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!

Answer

Give this a
try:

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