SQL Server 2008中两个字符串的返回差
有两个字符串。这些字符串可能有差异。我需要返回不同的值“Difference”和不同值的“Position” 上面的文章显示了一些类似的东西,但是,我的字符串没有任何分隔符,所以我在应用该方法时遇到了问题。两个字符串的长度始终为24个字符。但是,差异会有所不同,所以我不能只比较String1的位置1和String2的位置1 理想情况下,考虑到每个位置都有一个意义,表示方差值越少,表示相同的值越好。不过,让我展示一下这种差异会非常有帮助 这很难看,但是 首先,给自己弄一份。然后你可以这样做:SQL Server 2008中两个字符串的返回差,sql,sql-server-2008,ssms,Sql,Sql Server 2008,Ssms,有两个字符串。这些字符串可能有差异。我需要返回不同的值“Difference”和不同值的“Position” 上面的文章显示了一些类似的东西,但是,我的字符串没有任何分隔符,所以我在应用该方法时遇到了问题。两个字符串的长度始终为24个字符。但是,差异会有所不同,所以我不能只比较String1的位置1和String2的位置1 理想情况下,考虑到每个位置都有一个意义,表示方差值越少,表示相同的值越好。不过,让我展示一下这种差异会非常有帮助 这很难看,但是 首先,给自己弄一份。然后你可以这样做: D
DECLARE @String1 varchar(8000) = 'abcd10234619843ab13',
@String2 varchar(8000) = 'bbcd10234619843ac14';
WITH C AS(
SELECT @String1 AS String1,
@String2 AS String2,
S1.[position],
S1.token AS C1,
S2.token AS C2
FROM dbo.NGrams8k(@String1,1) S1
JOIN dbo.NGrams8k(@String2,1) S2 ON S1.[position] = S2.position)
SELECT (SELECT '' + C.C2
FROM C
WHERE C.C1 != C.C2
ORDER BY C.[position]
FOR XML PATH(''),TYPE).value('.','varchar(8000)') AS Difference,
(SELECT ISNULL(NULLIF(C.C2,C.C1),'-')
FROM C
ORDER BY C.[position]
FOR XML PATH(''),TYPE).value('.','varchar(8000)') AS Ideal,
STUFF((SELECT CONCAT(',',C.[position])
FROM C
WHERE C.C1 != C.C2
ORDER BY C.[position]
FOR XML PATH(''),TYPE).value('.','varchar(8000)'),1,1,'') AS Position;
如果您使用的是较新且受支持的版本的SQL Server,这实际上要容易得多,只需扫描一次所需的值:
DECLARE @String1 varchar(8000) = 'abcd10234619843ab13',
@String2 varchar(8000) = 'bbcd10234619843ac14';
WITH C AS(
SELECT @String1 AS String1,
@String2 AS String2,
S1.[position],
S1.token AS C1,
S2.token AS C2
FROM dbo.NGrams8k(@String1,1) S1
JOIN dbo.NGrams8k(@String2,1) S2 ON S1.[position] = S2.position)
SELECT STRING_AGG(NULLIF(C.C2,C.C1),'') WITHIN GROUP (ORDER BY C.position) AS Difference,
STRING_AGG(ISNULL(NULLIF(C.C2,C.C1),'-'),'') WITHIN GROUP (ORDER BY C.position) AS Ideal,
STRING_AGG(CASE C.C1 WHEN C.C2 THEN NULL ELSE C.[position] END,',') WITHIN GROUP (ORDER BY C.position) AS Position
FROM C
这很难看,但是
首先,给自己弄一份。然后你可以这样做:
DECLARE @String1 varchar(8000) = 'abcd10234619843ab13',
@String2 varchar(8000) = 'bbcd10234619843ac14';
WITH C AS(
SELECT @String1 AS String1,
@String2 AS String2,
S1.[position],
S1.token AS C1,
S2.token AS C2
FROM dbo.NGrams8k(@String1,1) S1
JOIN dbo.NGrams8k(@String2,1) S2 ON S1.[position] = S2.position)
SELECT (SELECT '' + C.C2
FROM C
WHERE C.C1 != C.C2
ORDER BY C.[position]
FOR XML PATH(''),TYPE).value('.','varchar(8000)') AS Difference,
(SELECT ISNULL(NULLIF(C.C2,C.C1),'-')
FROM C
ORDER BY C.[position]
FOR XML PATH(''),TYPE).value('.','varchar(8000)') AS Ideal,
STUFF((SELECT CONCAT(',',C.[position])
FROM C
WHERE C.C1 != C.C2
ORDER BY C.[position]
FOR XML PATH(''),TYPE).value('.','varchar(8000)'),1,1,'') AS Position;
如果您使用的是较新且受支持的版本的SQL Server,这实际上要容易得多,只需扫描一次所需的值:
DECLARE @String1 varchar(8000) = 'abcd10234619843ab13',
@String2 varchar(8000) = 'bbcd10234619843ac14';
WITH C AS(
SELECT @String1 AS String1,
@String2 AS String2,
S1.[position],
S1.token AS C1,
S2.token AS C2
FROM dbo.NGrams8k(@String1,1) S1
JOIN dbo.NGrams8k(@String2,1) S2 ON S1.[position] = S2.position)
SELECT STRING_AGG(NULLIF(C.C2,C.C1),'') WITHIN GROUP (ORDER BY C.position) AS Difference,
STRING_AGG(ISNULL(NULLIF(C.C2,C.C1),'-'),'') WITHIN GROUP (ORDER BY C.position) AS Ideal,
STRING_AGG(CASE C.C1 WHEN C.C2 THEN NULL ELSE C.[position] END,',') WITHIN GROUP (ORDER BY C.position) AS Position
FROM C
. . SQL并不是为这类事情而设计的。如果必须的话,可以使用递归CTE,但它会很混乱。SQL并不是为这类事情而设计的。如果必须的话,可以使用递归CTE,但它会很混乱。我会清理。工作起来很有魅力!我会打扫的。工作起来很有魅力!