awk - Extract time and source ip address from pcap file -

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I have a PAPT trace file and I want to remove time and source IP addresses from packets. I am using tcpdump and awk here is a sample of the file 

  02: 00: 00.001814 IP 61.31.228.1.80 & gt; 0.106.173.16.19999: Flag [S], Seac 4049606604, ACC 4044405336, 512 wins, length 0 02: 00: 00.005787 IP 61.31.228.1.80 & gt; 0.4.173.19.13923: Flag [S], Seek 3812128115, AK 3811406374, 512 wins, length 0 02: 00: 00.005799 IP 74.54.182.242.80 & gt; 0.176.229.61.43527: Flags [S], SAC 61247722, AK 352633207, WIN 65535, OPTION [MSC 1460, NOP, NOP, SAPOK], length 0   
 Then I applied the CSV To remove time and source IP address in the file, awk:  
  02: 00: 00.001814,0.106.173.16.199 99: 02: 00: 00.005787,0.4.173.19.13923: 02: 00: 00.005799,0.176.229.61.43527:   
 I am only interested in the last bit of time and I am getting rid of "" at the end of the source IP ".   
 
 
 
 
 
 
 
 
 
 
 
 Div class = "post-text" itemprop = "text"> 
 Try this:  
  awk -F '[:.] "$ 5 ==" IP "{print $ 4 "$ 12", "$ 13". "$ 14". "$ 15} '  
 Output  
  001814,0.106.173.16 005787 , 0.4.173.19 005799,0.176.229.61

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