The question above is all I need help with. I have two common separated lists and I need to compare the two lists and return the values that are in both lists.
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1Normalize the data properly, then just use an `INNER JOIN`. – David Jan 26 '16 at 15:46
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Agree with @David. I take it you have a list of values in a single field, in your SQL database. That is not proper normalized form. You need to put the values in a separate table. Once you have a properly normalized db, then you can expect SQL to do the comparison for you. – Jan 26 '16 at 15:51
2 Answers
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You could, for instance, parse each list into a table (variable) and then use an inner join to obtain the common values.

cdonner
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You can use the Split
function from this SO post and then get the common elements using INTERSECT
:
DECLARE @String1 VARCHAR(50)='A,B,C'
DECLARE @String2 VARCHAR(50)='A,B'
SELECT * FROM dbo.Split(@String1, ',')
INTERSECT
SELECT * FROM dbo.Split(@String2, ',')
I am referring to this Split function as originally posted by Romil Kumar Jain here.
CREATE FUNCTION Split (
@InputString VARCHAR(8000),
@Delimiter VARCHAR(50)
)
RETURNS @Items TABLE (
Item VARCHAR(8000)
)
AS
BEGIN
IF @Delimiter = ' '
BEGIN
SET @Delimiter = ','
SET @InputString = REPLACE(@InputString, ' ', @Delimiter)
END
IF (@Delimiter IS NULL OR @Delimiter = '')
SET @Delimiter = ','
--INSERT INTO @Items VALUES (@Delimiter) -- Diagnostic
--INSERT INTO @Items VALUES (@InputString) -- Diagnostic
DECLARE @Item VARCHAR(8000)
DECLARE @ItemList VARCHAR(8000)
DECLARE @DelimIndex INT
SET @ItemList = @InputString
SET @DelimIndex = CHARINDEX(@Delimiter, @ItemList, 0)
WHILE (@DelimIndex != 0)
BEGIN
SET @Item = SUBSTRING(@ItemList, 0, @DelimIndex)
INSERT INTO @Items VALUES (@Item)
-- Set @ItemList = @ItemList minus one less item
SET @ItemList = SUBSTRING(@ItemList, @DelimIndex+1, LEN(@ItemList)-@DelimIndex)
SET @DelimIndex = CHARINDEX(@Delimiter, @ItemList, 0)
END -- End WHILE
IF @Item IS NOT NULL -- At least one delimiter was encountered in @InputString
BEGIN
SET @Item = @ItemList
INSERT INTO @Items VALUES (@Item)
END
-- No delimiters were encountered in @InputString, so just return @InputString
ELSE INSERT INTO @Items VALUES (@InputString)
RETURN
END -- End Function
GO
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The strings have already been created and are stored in a variable... How would you then manipulate the code to solve the problem. – TheSQLLearner Jan 26 '16 at 15:56
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Updated post, strings are now stored in variables called `@String1` and `@String2` – Alex Jan 26 '16 at 16:03
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All I could find on this website was how to return true, false, 1 or 0 if values matched... Not the actual values themselves. – TheSQLLearner Jan 26 '16 at 16:22