2字符串Server 2008的返回差

Question

有两个字符串。这些字符串可能有差异。我需要返回不同价值观的“差异”和不同价值观的“位置”。

How to find diff between two string in SQL

上面的帖子显示了一些类似的内容，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困难。这两个字符串将始终是24个字符长。但是，差异将是不同的，所以我不能仅仅将String1的位置1与String2的位置1进行比较。

理想情况下，考虑到每个位置都有意义，表示方差值减去任何相同的值会更好。然而，让我仅仅显示出不同之处会非常有帮助。

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Answer 1

这很难看但是..。

首先，给自己买一份NGrams8K的副本。然后你可以做这样的事情：

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;

如果您使用的是最近的版本，并且支持Server的版本，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