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I am trying to parse dhcp.lease file from /tmp/dhcp.lease in openwrt . 

root@OpenWrt:/# cat /tmp/dhcp.leases

1568953482  70:B3:D5:14:D0:31 192.168.3.51 device1 01:70:B3:D5:14:D0:31
2867821468  38:B8:EB:10:00:22 192.168.5.93 device2 01:38:B8:EB:10:00:22
8984532872  00:01:0A:33:11:33 192.168.5.44 CISCOee 01:00:01:0A:33:11:33

Where, 2nd column - MAC address , 3rd column - IP address and 4th column- Name

I want to run a shell script to parse No of devices , MAC address , IP address and Device Name from this dhcp.lease list .

sample output :

if there is 3 device list is present in dhpcp.lease file , I want to print output like :

3
70:B3:D5:14:D0:31/192.168.3.51/device1
38:B8:EB:10:00:22/192.168.5.93/device2
00:01:0A:33:11:33/192.168.5.44/CISCOee

and if no device list found , it should return 

0

Can i do with simple file content iteration ? or any fast method like sed/awk ? Any sample code ?

vkp_stack
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3 Answers3

1

awk command may help you.

$ awk -v OFS='\n' '$2 ~ /[0-9A-Z]:/{n=n+1;a[n]=$2"/"$3"/"$4} END{print n==""?0:n; for(i in a)print a[i]}' /tmp/dhcp.leases

Brief explanation,

  • print the line which $2 matched to regex [0-9A-Z]:
  • save the count of match to n, and also save $2/$3/$3 to the array a
  • print the count of matched case n and the value in the array a in the end
CWLiu
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1

The previous awk statement seems over complicated to me. The following is simpler:

awk '{ dat[NR]=$2"/"$3"/"$4 } END { print NR;for ( i=1;i<=NR;i++) { print dat[i] } }'  /tmp/dhcp.lease

The leases file is in a fixed format with just the data required and the total count will be the number of records (NR variable). Read the data required into an array dat and then print NR and loop through the array printing each entry.

Raman Sailopal
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0

Besides awk, you can use grep with regex and using a temporary file in where you'll save the matched output for further counting its lines and displaying it like this example:

~$ cat /tmp/dhcp.leases | grep -oE "([0-9A-Z]{2}:){5}[0-9A-Z]{2}.*([0-9]{1,3}\.){3}[0-9]{1,3}.*[a-zA-Z]{3,}[0-9]+?" | tr ' ' '/' > /tmp/testFile
~$ cat /tmp/testFile | wc -l; cat /tmp/testFile

Output:

3
70:B3:D5:14:D0:31/192.168.3.51/device1
38:B8:EB:10:00:22/192.168.5.93/device2
00:01:0A:33:11:33/192.168.5.44/CISCOee

Explanation:

  • grep -oE: Print only the matched (non-empty) parts of the matching line and use the Extended regexp see the man page.
  • Redirect the output to a temprary file /tmp/testFile.
  • Count the lines of /tmp/testFile using wc and print its lines.
Chiheb Nexus
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