SQL - Find Records Where Values Match Characters in Comma Delimited String

I am using a PATINDEX in a query to find values within a comma delimited string.  However, if a string is holding values (Z, A), the query only returns those records where both Z and A are contained in the string and in that order.  What I need is a SQL function that would find records that contain both the Z and the A (Z , A) or A and Z (A , Z) or only Z or only A.  The wild cards % does not seems to help find records where the code is a single letter and does not meet both criteria.


DECLARE @long1 decimal(5,2); 
DECLARE @lat1 decimal(5,2);
DECLARE @rangeFactor decimal(7,6);
SET @rangeFactor = 0.014457;
SELECT @lat1 = LATITUDE, @long1 = LONGITUDE
FROM dbo.zipcodes where zipcode = '" + Replace(rsC9__varZ2, "'", "''") + "';
SELECT DISTINCT(C.CommunityName), B.ABBR, B.Latitude, B.Longitude, B.zipcode, C.CommunityID, C.CommCity, C.CommState, C.CommZip, CommCode1         
FROM dbo.zipcodes B inner join dbo.Communities C ON B.ZipCode = C.CommZip
WHERE B.LATITUDE BETWEEN @lat1-(100 *@rangeFactor) and @lat1+(100 *@rangeFactor)
AND B.LONGITUDE BETWEEN @long1-(100 *@rangeFactor) and @long1+(100 *@rangeFactor)
AND dbo.getDistance(@lat1,@long1,B.latitude,B.longitude) <= 100
AND (PATINDEX('%" + Replace(rsC9__varS4, "'", "''") + "%', CommCode1) > 0)
ORDER BY C.CommunityName ASC"

Open in new window

LVL 1
CraigDeringtonAsked:
Who is Participating?

[Product update] Infrastructure Analysis Tool is now available with Business Accounts.Learn More

x
I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

reb73Commented:
Create an UDF that parses delimited strings to a table like the one in the link below -

http://www.experts-exchange.com/Microsoft/Development/MS-SQL-Server/Q_22819980.html?sfQueryTermInfo=1+10+dbo.fnsplit

Then just do an inner join on this UDF like the template SQL code below (code from line 7 to 12 in your code being replaced with the code below)
SELECT DISTINCT(C.CommunityName), B.ABBR, B.Latitude, B.Longitude, B.zipcode, C.CommunityID, C.CommCity, C.CommState, C.CommZip, CommCode1         
FROM dbo.zipcodes B inner join dbo.Communities C ON B.ZipCode = C.CommZip
inner join dbo.fnsplit(' + REPLACE(rsC9__varS4, "'", "''") + ', ',') F ON C.CommCode1 LIKE ''%'' + F.SplitCol + ''%''
WHERE B.LATITUDE BETWEEN @lat1-(100 *@rangeFactor) and @lat1+(100 *@rangeFactor)
AND B.LONGITUDE BETWEEN @long1-(100 *@rangeFactor) and @long1+(100 *@rangeFactor)
AND dbo.getDistance(@lat1,@long1,B.latitude,B.longitude) <= 100
ORDER BY C.CommunityName ASC"

Open in new window

0
8080_DiverCommented:
The provided code looks somewhat like a stored procedure or a portion thereof.  What you need to do is to break the comma separated list of characters into something that will work (perhaps an in-memory table) and then build your SQL query so that it makes use of that something.  
I am assuming that rsC9__varS4 is an input variable to the stored procedure.  So, the following code should be something like what you will need.
By the way, DISTINCT doesn't work on a column by column basis but on a row by row basis, so you DISTINCT(C.CommunityName) isn't really correct syntax. ;-)
 

DECLARE @long1 decimal(5,2); 
DECLARE @lat1 decimal(5,2);
DECLARE @rangeFactor decimal(7,6);
SET @rangeFactor = 0.014457;
SELECT @lat1 = LATITUDE, 
       @long1 = LONGITUDE
FROM   dbo.zipcodes 
where zipcode = '" + Replace(rsC9__varZ2, "'", "''") + "';
 
DECLARE @TempListTable TABLE
                       (
                        ACharacter AS VarChar(10)
                       );
 
DECLARE @TempChars AS VarChar(10);
DECLARE @WorkString AS VarChar(8000);
DECLARE @CommaPos AS INT;
 
-- Set the WorkString variable to an empty string
SET @WorkString = '';
 
-- Find the position of the first comma, if any
SET @CommaPos	=	CharIndex(',', rsC9__varS4);
 
IF (@CommaPos > 0)
THEN
BEGIN
-- We got a comma, so add the first string and set the WorkString to the rest
	INSERT LEFT(rsC9__varS4, @CommaPos - 1) INTO @TempListTable ;
	SET @WorkString = RIGHT(rsC9__varS4, LENGTH(rsC9__varS4) - @CommaPos);
END
 
WHILE (@WorkString <> '')
BEGIN
	SET @CommaPos	=	CharIndex(',', @WorkString);
 
	IF (@CommaPos > 0)
	THEN
	BEGIN
	-- We got a comma, so add the first string and set the WorkString to the rest
		INSERT LEFT(@WorkString, @CommaPos - 1) INTO @TempListTable ;
		SET @WorkString = RIGHT(@WorkString, LENGTH(@WorkString) - @CommaPos);
	END
 
END
 
 
SELECT DISTINCT 
       C.CommunityName, 
       B.ABBR, 
       B.Latitude, 
       B.Longitude, 
       B.zipcode, 
       C.CommunityID, 
       C.CommCity, 
       C.CommState, 
       C.CommZip, 
       CommCode1         
FROM   dbo.zipcodes B 
inner join dbo.Communities C 
  ON   B.ZipCode = C.CommZip
JOIN   @TempListTable  Z
WHERE  B.LATITUDE BETWEEN @lat1-(100 *@rangeFactor) and @lat1+(100 *@rangeFactor)
  AND B.LONGITUDE BETWEEN @long1-(100 *@rangeFactor) and @long1+(100 *@rangeFactor)
  AND dbo.getDistance(@lat1,@long1,B.latitude,B.longitude) <= 100
  AND (PATINDEX('%' + Z.ACharacter  + '%', CommCode1) > 0)
ORDER BY C.CommunityName ASC"

Open in new window

0

Experts Exchange Solution brought to you by

Your issues matter to us.

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Start your 7-day free trial
CraigDeringtonAuthor Commented:
Thanks for your help with this.  Your solution worked great...
0
CraigDeringtonAuthor Commented:
Excellent.  Thank you.
0
It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today
Microsoft SQL Server 2005

From novice to tech pro — start learning today.