考虑下面的带有六个GROUP BY 子查询的MS Access 查询,它确实显示了哪些项目在所有列中匹配。但对于其他列的差异,查询提供了一种仪表板供用户决定匹配。
查询背后的概念是将所有分数连接成一个字符串(例如,11211、1111、01111),然后逐个字符检查(使用Left和Mid()字符串函数)项目之间的字符数是多少大于一。此外,对于多次重复出现,我必须使用 @。
SELECT t1.Item,
(t1.Score1 & t1.Score2 & t1.Score3 & t1.Score4 & t1.Score5) AS StringScores,
(SELECT 'Yes'
FROM Scores t2
WHERE (t1.Score1 & t1.Score2 & t1.Score3 & t1.Score4 & t1.Score5)=
(t2.Score1 & t2.Score2 & t2.Score3 & t2.Score4 & t2.Score5)
GROUP BY 'Yes', Cstr(t2.Score1 & t2.Score2 & t2.Score3 & t2.Score4 & t2.Score5)
HAVING Count(*) > 1) AS [All Five Scores Match ?],
(SELECT 'Yes @ ' & Left((t3.Score1 & t3.Score2 & t3.Score3 & t3.Score4 & t3.Score5), 1)
FROM Scores t3
WHERE Left((t1.Score1 & t1.Score2 & t1.Score3 & t1.Score4 & t1.Score5), 1)=
Left((t3.Score1 & t3.Score2 & t3.Score3 & t3.Score4 & t3.Score5), 1)
GROUP BY 'Yes @ ' & Left((t3.Score1 & t3.Score2 & t3.Score3 & t3.Score4 & t3.Score5), 1)
HAVING Count(*) > 1) AS [First Score Matches ?],
(SELECT 'Yes @ ' & Mid((t4.Score1 & t4.Score2 & t4.Score3 & t4.Score4 & t4.Score5), 2, 1)
FROM Scores t4
WHERE Mid((t1.Score1 & t1.Score2 & t1.Score3 & t1.Score4 & t1.Score5), 2, 1)=
Mid((t4.Score1 & t4.Score2 & t4.Score3 & t4.Score4 & t4.Score5), 2, 1)
GROUP BY 'Yes @ ' & Mid((t4.Score1 & t4.Score2 & t4.Score3 & t4.Score4 & t4.Score5), 2, 1)
HAVING Count(*) > 1) AS [Second Score Matches ?],
(SELECT 'Yes @ ' & Mid((t5.Score1 & t5.Score2 & t5.Score3 & t5.Score4 & t5.Score5), 3, 1)
FROM Scores t5
WHERE Mid((t1.Score1 & t1.Score2 & t1.Score3 & t1.Score4 & t1.Score5), 3, 1)=
Mid((t5.Score1 & t5.Score2 & t5.Score3 & t5.Score4 & t5.Score5), 3, 1)
GROUP BY 'Yes @ ' & Mid((t5.Score1 & t5.Score2 & t5.Score3 & t5.Score4 & t5.Score5), 3, 1)
HAVING Count(*) > 1) AS [Third Score Matches ?],
(SELECT 'Yes @ ' & Mid((t6.Score1 & t6.Score2 & t6.Score3 & t6.Score4 & t6.Score5), 4, 1)
FROM Scores t6
WHERE Mid((t1.Score1 & t1.Score2 & t1.Score3 & t1.Score4 & t1.Score5), 4, 1)=
Mid((t6.Score1 & t6.Score2 & t6.Score3 & t6.Score4 & t6.Score5), 4, 1)
GROUP BY 'Yes @ ' & Mid((t6.Score1 & t6.Score2 & t6.Score3 & t6.Score4 & t6.Score5), 4, 1)
HAVING Count(*) > 1) AS [Fourth Score Matches ?],
(SELECT 'Yes @ ' & Mid((t7.Score1 & t7.Score2 & t7.Score3 & t7.Score4 & t7.Score5), 5, 1)
FROM Scores t7
WHERE Mid((t1.Score1 & t1.Score2 & t1.Score3 & t1.Score4 & t1.Score5), 5, 1) =
Mid((t7.Score1 & t7.Score2 & t7.Score3 & t7.Score4 & t7.Score5), 5, 1)
GROUP BY 'Yes @ ' & Mid((t7.Score1 & t7.Score2 & t7.Score3 & t7.Score4 & t7.Score5), 5, 1)
HAVING Count(*) > 1) AS [Fifth Score Matches ?]
FROM Scores AS t1;
下面是输出。如您所见,用户可以通过第一列(主要内容)告诉 A 和 B 在所有 5 个分数中完全匹配;和 C & D 匹配所有四个分数,除了第一个; C 匹配前两个项目 A 和 B,但第三个分数除外; D 在除第一和第三得分之外的所有得分上都匹配前两个 A 和 B。
我继续测试了另外四个可能的分数:
诚然,这些输出可能看起来像是对原始表格的重新样式化,但请记住,只会出现大于 1 的出现次数。您可以将此查询输出到表中并过滤/排序以更清晰地查看模式。
最后,如果您需要检查前三个分数、后两个分数或多个分数的任意组合,则可以扩展此功能。只需使用Left(StringScores, 2) 或Mid(StringScores, 4, 2) 字符串函数添加相应的子查询。