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Just to point out the obvious, you've got a closing bracket missing after the "(?!a" in the two examples you posted. I'm assuming it is a typing error, but thought I'd better check:

Regexp_substr( 'B2B' , '(?<!a)\d(?!a' )

Regexp_replace( 'B2B' , '(?<!a)\d(?!a', '#' )


-Paul.


-----Original Message-----
From: WDSCI-L [mailto:wdsci-l-bounces@xxxxxxxxxxxx] On Behalf Of Craig Richards
Sent: 28 April 2016 10:42
To: Rational Developer for IBM i / Websphere Development Studio Client for System i & iSeries
Subject: [WDSCI-L] DB2 regex bug with look around ?

Apologies if this isn't quite the correct forum for this question, but it seems a good place to get a chance of a second opinion...

Has anyone noticed anything funny with the look around regex capabilities of DB2 ( 7.1 )

For example

Applying

(?<!a)\d(?!a)

To

B2B

It should return

2

But it seems to find a match but with no text returned.

E.g if you run:

Select
Regexp_substr( 'B2B' , '(?<!a)\d(?!a' )
From sysibm/sysdummy1

It doesn't return null and it appears as blank in STRSQL.

whereas if you run the regex over 'a2B' you clearly get null returned.

But it should select the digit as far as I understand.

Further,

Select
Regexp_replace( 'B2B' , '(?<!a)\d(?!a', '#' ) From sysibm/sysdummy1

Will return B#2B

which seems to indicate that the regex selected the position just after the B where the negative look behind matched, but then didn't appear to go on and match the digit.

Does this make sense to anyone?

The regex is saying "find me a digit which doesn't have an "a" on either side of it"

Am I missing something obvious?

Thanks,
Craig
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