Re: Need regexp help (seemingly simple problem)
Date: Fri, 23 Oct 2009 12:10:32 -0700
The [:digit:] class actually need to be written as [[:digit:]].
Here's an example using regexp_replace
select '2B4a12345Z2lr7' source
- want from 'a' to 'Z' , regexp_replace('2B4a12345Z2lr7','^.*a([[:digit:]]+)Z.*$','\1',1,1) target , regexp_replace('2B4a12345Z2lr7','^.*a(+)Z.*$','\1',1,1) target from dual /
12:09:39 ordevdb01.radisys.com - js001292_at_dv11 SQL> /
SOURCE TARGE TARGE
-------------- ----- -----
2B4a12345Z2lr7 12345 12345
1 row selected.
On Fri, Oct 23, 2009 at 11:01 AM, Michael Moore <michaeljmoore_at_gmail.com>wrote:
> First of all I would like to say that finding detailed information about
> Oracle's implementation of regular expressions is next to impossible. Either
> that, or I am looking in all the wrong places. So, links to any complete
> online documentation would be appreciated. I found plenty of 'basic
> explanation' but there has got to be more.
> Here is the specific problem I am trying to solve using REGEXP_SUBSTR. I
> have a string that might look something like this:
> '2B4a12345Z2lr7' . I want to extract the 12345. In words, find the first
> letter 'a' and return all the digits up to the letter Z.
> There can be any number of random non-'a' characters prior to the initial
> 'a' and likewise after the letter Z. In other words, in my example, '2B4'
> just represents some random sequence of characters, as does '2lr7'.
> I could easily use INSTR and SUBSTR to do this, but my goal is to solve the
> problem using a single REGEXP_SUBSTR statement. Also, I have used the
> pattern 'a[[:digit:]]+' which results in 'a12345'. This is almost right, but
> I don't want the 'a' to be returned.