SELECT with a Replace()

Question

I have a table of names and addresses, which includes a postcode column. I want to strip the spaces from the postcodes and select any that match a particular pattern. I\'m tryin

Answer 1

if you want any hope of ever using an index, store the data in a consistent manner (with the spaces removed). Either just remove the spaces or add a persisted computed column, Then you can just select from that column and not have to add all the space removing overhead every time you run your query.

add a PERSISTED computed column:

ALTER TABLE Contacts ADD PostcodeSpaceFree AS Replace(Postcode, ' ', '') PERSISTED 
go
CREATE NONCLUSTERED INDEX IX_Contacts_PostcodeSpaceFree 
ON Contacts (PostcodeSpaceFree) --INCLUDE (covered columns here!!)
go

to just fix the column by removing the spaces use:

UPDATE Contacts
    SET Postcode=Replace(Postcode, ' ', '')

now you can search like this, either select can use an index:

--search the PERSISTED computed column
SELECT 
    PostcodeSpaceFree 
    FROM Contacts
    WHERE PostcodeSpaceFree  LIKE 'NW101%'

or

--search the fixed (spaces removed column)
SELECT 
    Postcode
    FROM Contacts
    WHERE PostcodeLIKE 'NW101%'

Answer 2

SELECT *
FROM Contacts
WHERE ContactId IN
    (SELECT a.ContactID
    FROM
        (SELECT ContactId, Replace(Postcode, ' ', '') AS P
        FROM Contacts
        WHERE Postcode LIKE '%N%W%1%0%1%') a
    WHERE a.P LIKE 'NW101%')