क्या आप यही खोज रहे हैं?
सही / गलत परिणाम
-- matches only those values which exist in both CSV sets
SELECT T1.[Item], CASE WHEN T2.[Item] IS NULL THEN 0 ELSE 1 END AS [Match]
FROM [dbo].[Split]('val1,val2,val3', ',') AS T1
LEFT JOIN [dbo].[Split]('val3,val4', ',') AS T2 on T1.[Item] = T2.[Item]
रिटर्न
Item Match
val1 0
val2 0
val3 1
केवल सच्चे मिलान
-- matches only those values which exist in both CSV sets
SELECT T1.[Item]
FROM [dbo].[Split]('val1,val2,val3', ',') AS T1
INNER JOIN [dbo].[Split]('val3,val4', ',') AS T2 on T1.[Item] = T2.[Item]
रिटर्न
Item
val3
विभाजन समारोह
CREATE FUNCTION [dbo].[Split]
(
@s VARCHAR(max),
@split CHAR(1)
)
RETURNS @temptable TABLE ([Item] VARCHAR(MAX))
AS
BEGIN
DECLARE @x XML
SELECT @x = CONVERT(xml,'<root><s>' + REPLACE(@s,@split,'</s><s>') + '</s></root>');
INSERT INTO @temptable
SELECT [Value] = T.c.value('.','varchar(20)')
FROM @X.nodes('/root/s') T(c);
RETURN
END;