Обсуждение: LIKE Command

Hi,

I have just upgraded to 7.2 and found a bug in the LIKE command. The
command I am using is

SELECT AccNum , StuID , StuInsUID , RefPhyNam , StuDes , PatAge , PatSiz
, PatWei , Owner , GroupName , Priv , PatParent , StuDat , StuTim ,
InsertDate , InsertTime , NumStuRelSer , NumStuRelIma FROM StudyLevel
WHERE StuInsUID like '1.2.826.0.1.4156771.2155605229.6692.1020158785.1'
AND PatParent = 'BONES';

This previously worked but now it will only work if I put some sort of
expression in the string (ie a % at the end).


Is this intended now or is it a bug.

Thanks in advance
--
Regards

Bob
_______________________________________________________________________
Bob Kenny,
Link Medical Limited,
Unit 9,
Moor Place Farm,
Plough Lane
Bramshill,
Hampshire RG27 0RF
England

Tel : (+44) 0118 9326850
Fax : (+44) 0118 9326717

Email : bob@linkmed.org
Mobile: (+44) 7711 691880
_______________________________________________________________________

Bob Kenny <bob@linkmed.org> writes:
> I have just upgraded to 7.2 and found a bug in the LIKE command.

You're going to have to provide a complete example to convince us that
LIKE is broken.

            regards, tom lane

...
> This previously worked but now it will only work if I put some sort of
> expression in the string (ie a % at the end)...

I am guessing that you have some blanks or other characters at the end
of the string. Simple test cases here do not show a problem.

                     - Thomas