本文介绍了一种在网上难以找到的有效使用 ACCESS 中 COUNTDISTINCT 方法的技巧。通过一段 SQL 查询示例，解决了如何准确地统计不重复记录的问题。
网上查了很多关于ACCESS的COUNT DISTINCT的方法，都不好使，在一个老外的贴子上找到了方法，贴出来
SELECT C.immu, T2.N
FROM [SELECT T.immu, Count(T.immu) AS N
FROM
(SELECT DISTINCT immu, Count(*) AS N
FROM plan GROUP BY immu) AS T
GROUP BY T.immu]. AS T2 INNER JOIN plan AS C ON T2.immu = C.immu
GROUP BY C.immu, T2.N;
SELECT C.immu, T2.N
FROM [SELECT T.immu, Count(T.immu) AS N
FROM
(SELECT DISTINCT immu, Count(*) AS N
FROM plan GROUP BY immu) AS T
GROUP BY T.immu]. AS T2 INNER JOIN plan AS C ON T2.immu = C.immu
GROUP BY C.immu, T2.N;
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