Oracle-L: RE: Sql question : use of SUBSTR/INSTR functions

From: Fink, Dan <Dan.Fink_at_mdx.com>
Date: Wed, 16 Oct 2002 13:05:43 -0800
Message-ID: <F001.004EB4D1.20021016130543@fatcity.com>

Try this. It uses the INSTR function to determine the start and end of the SUBSTR. 1 select substr('333.22.1.000',1,instr('333.22.1.000','.')-1) octet1,

  2         substr('333.22.1.000',
  3                (instr('333.22.1.000','.',1,1) + 1),
  4                (instr('333.22.1.000','.',1,2) -
instr('333.22.1.000','.',1,1))-1) octect2,
  5         substr('333.22.1.000',
  6                (instr('333.22.1.000','.',1,2) + 1),
  7                (instr('333.22.1.000','.',1,3) -
instr('333.22.1.000','.',1,2))-1) octect3,
  8         substr('333.22.1.000',(instr('333.22.1.000','.',1,3) + 1))

octect3
9* from dual
SQL> / OCT OC O OCT
--- -- - ---
333 22 1 000

Dan Fink

-----Original Message-----
Sent: Wednesday, October 16, 2002 2:32 PM To: Multiple recipients of list ORACLE-L

Help!

Anybody have a quick and dirty to parse the 4 octets of a typical IP address into 4 separate values. I will insert these into a table where database checks may verify that the data is in fact a number and also part of a valid ip range (the second thru fourth octets cannot be higher than 255. The source data is very dirty and often fat-fingered, hence the painful solution):

e.g.: 127.0.0.1 into 127 (val 1), 0 (val 2), 0 (val 3) and 1 (val 4).

I have used various flavors of substr/instr to unravel this, but the varying length of the octets (up to 3 bytes) defeats my rudimentary sql coding skills. I probably have to attack the IP with decode, and any input will be very welcome.

Running V 8.1.6.

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Author: Fink, Dan
INET: Dan.Fink_at_mdx.com

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