Return 2 个字符串的差异 SQL Server 2008
Return Difference of 2 Strings SQL Server 2008
有2个字符串。这些字符串可能有差异。我需要 return 不同的值 "Difference" 和那些不同的值的 "Position"。
上面的 post 显示了类似的内容,但是,我的字符串没有任何分隔,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意义,表示方差值减去任何相同的值会更好。但是,让我展示差异会非常有帮助。
这很丑,但是...
首先,给自己买一份 NGrams8K
。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 支持的 版本的 SQL 服务器,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
有2个字符串。这些字符串可能有差异。我需要 return 不同的值 "Difference" 和那些不同的值的 "Position"。
上面的 post 显示了类似的内容,但是,我的字符串没有任何分隔,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意义,表示方差值减去任何相同的值会更好。但是,让我展示差异会非常有帮助。
这很丑,但是...
首先,给自己买一份 NGrams8K
。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 支持的 版本的 SQL 服务器,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