[SOLVED] Parsing datetime domain names

hattori.hanzo · 10-25-2010, 01:15 AM

I have spent lots of time trying to figure out how to parse the domain part of this file but am stuck. How should this be approached:

Code:

20100618,00:21:31,aaa.bbb.ccc.ddd,A,(6):12003(4)wins(4)domain(6)dsmain(3)com(0)
20100618,00:21:32,aaa.bbb.ccc.ddd,SOA,(12)SP-CR-S003(4)emea(6)domain(3)com(0)
20100618,00:21:33,aaa.bbb.ccc.ddd,A,(7)altfarm(9)mediaplex(3)com(0)
20100618,00:21:33,aaa.bbb.ccc.ddd,A,(3)img(9)mediaplex(3)com(0)

Code:

2010-06-18 00:21:31,aaa.bbb.ccc.ddd,A,com,dsmain,domain,wins,:12003
2010-06-18 00:21:32,aaa.bbb.ccc.ddd,SOA,com,domain,emea,SP-CR-S003,SOA
2010-06-18 00:21:33,aaa.bbb.ccc.ddd,A,com,mediaflex,altfarm
20100-6-18 00:21:33,aaa.bbb.ccc.ddd,A,com,mediaflex,altfarm

The domains need to be parse at a maximum for 4 levels then the remainder will be put the last field.

Thanks & Regards,

druuna · 10-25-2010, 03:46 AM

Hi,

Is this what you are looking for:

sed 's/([0-9]*)/@/g' input | awk -F, '{ printf("%s,%s,%s,%s", $1, $2, $3, $4) ; z=split($5,tets,"@") ; for (i = z-1; i > 1; --i) printf(",%s", tets[i]) ; print "" }'

Example run on given example:

Code:

$ cat input
20100618,00:21:31,aaa.bbb.ccc.ddd,A,(6):12003(4)wins(4)domain(6)dsmain(3)com(0)
20100618,00:21:32,aaa.bbb.ccc.ddd,SOA,(12)SP-CR-S003(4)emea(6)domain(3)com(0)
20100618,00:21:33,aaa.bbb.ccc.ddd,A,(7)altfarm(9)mediaplex(3)com(0)
20100618,00:21:33,aaa.bbb.ccc.ddd,A,(3)img(9)mediaplex(3)com(0)

$ sed 's/([0-9]*)/@/g' input | awk -F, '{ printf("%s,%s,%s,%s", $1, $2, $3, $4) ; z=split($5,tets,"@") ; for (i = z-1; i > 1; --i) printf(",%s", tets[i]) ; print "" }'
20100618,00:21:31,aaa.bbb.ccc.ddd,A,com,dsmain,domain,wins,:12003
20100618,00:21:32,aaa.bbb.ccc.ddd,SOA,com,domain,emea,SP-CR-S003
20100618,00:21:33,aaa.bbb.ccc.ddd,A,com,mediaplex,altfarm
20100618,00:21:33,aaa.bbb.ccc.ddd,A,com,mediaplex,img

Hope this helps.

EDIT
I just noticed you also want to change the date itself (I focussed on the reversing of the domain parts). Here's a solution (long one-liner...) that does both:

sed -e 's/^$[0-9]\{4\}$$[0-9]\{2\}$$[0-9]\{2\}$,/\1-\2-\3 /' -e 's/([0-9]*)/@/g' input | awk -F, '{ printf("%s,%s,%s", $1, $2, $3) ; z=split($4,tets,"@") ; for (i = z-1; i > 1; --i) printf(",%s", tets[i]) ; print "" }'

Example run:

Code:

$ sed -e 's/^\([0-9]\{4\}\)\([0-9]\{2\}\)\([0-9]\{2\}\),/\1-\2-\3 /' -e 's/([0-9]*)/@/g' input | awk -F, '{ printf("%s,%s,%s", $1, $2, $3) ; z=split($4,tets,"@") ; for (i = z-1; i > 1; --i) printf(",%s", tets[i]) ; print "" }'
2010-06-18 00:21:31,aaa.bbb.ccc.ddd,A,com,dsmain,domain,wins,:12003
2010-06-18 00:21:32,aaa.bbb.ccc.ddd,SOA,com,domain,emea,SP-CR-S003
2010-06-18 00:21:33,aaa.bbb.ccc.ddd,A,com,mediaplex,altfarm
2010-06-18 00:21:33,aaa.bbb.ccc.ddd,A,com,mediaplex,img

/EDIT

hattori.hanzo · 10-25-2010, 09:27 AM

This works perfectly. Thank you very much.

Your coding is an inspiration to me to better my skills and also to learn from.

druuna · 10-25-2010, 09:31 AM

Hi,

You're welcome

Quote:

Your coding is an inspiration to me to better my skills and also to learn from.

If anything is unclear about the one-liner I provided I will gladly explain it to you.