如何在 MS-Access 中计算一个 ID 上的三个不同的不同值和组？

Question

所以我知道 MS-Access 不允许 SELECT COUNT(DISTINCT....) FROM ...，但我正在尝试找到一个更可行的替代通常标准的替代方法

SELECT COUNT(*) FROM (SELECT DISTINCT Name FROM table1)

我的问题是我正在尝试执行三个单独的 Count 函数并将它们按 ID 分组。如果我使用上面的方法，它会给我整个表的唯一值总数，而不是仅 ID 值的总数。我试过做

(SELECT COUNT(*) FROM (SELECT DISTINCT Name FROM table1 as T2
WHERE T2.ColumnA = T1.ColumnA)) As MyVal
FROM table1 as T1

但它告诉我需要为 T1.ColumnA 指定一个值。

我要完成的 SQL 查询是这样的：

SELECT ID
COUNT(DISTINCT ColumnA) as CA,
COUNT(DISTINCT ColumnB) as CB,
COUNT(DISTINCT ColumnC) as CC
FROM table1
GROUP BY ID

有什么想法吗？

Answer 1

您可以使用子查询。假设您有一个表，其中每个 id 出现一次：

select (select count(*)
        from (select columnA
              from table1 t1
              where t1.id = t.id
              group by columnA
             ) as a
       ) as num_a,
       (select count(*)
        from (select columnB
              from table1 t1
              where t1.id = t.id
              group by columnB
             ) as b
       ) as num_b,
       (select count(*)
        from (select columnC
              from table1 t1
              where t1.id = t.id
              group by columnC
             ) as c
       ) as num_c
from <table with ids> as t;

我不确定你是否认为这是“可行的”。

编辑：

这使它变得更加复杂。 . .它表明 MS Access 不支持超过一级深度的关联子句（您是否考虑切换到另一个数据库？）。

无论如何，蛮力的方式：

select a.id, a.numA, b.numB, c.numC
from ((select id, count(*) as numA
       from (select id, columnA
             from table1 t1
             group by id, columnA
            ) as a
      ) as a inner join
      (select id, count(*) as numB
       from (select id, columnB
             from table1 t1
             group by id, columnB
            ) as b
      ) as b
      on a.id = b.id
     ) inner join
     (select id, count(*) as numC
      from (select id, columnC
            from table1 t1
            group by id, columnC
           ) as c
     ) c
     on c.id = a.id;