Question

有关计算SQL中两个表的唯一字段组合的问题。 MSaccess 2003。

table1: id, fld1, fld2
table2: id, dateAndTime

id不是唯一的;表格在“id”之间加入。

我需要按fld1和fld2（int）的不同组合按日期（dd / mm / yy - 忽略时间）进行计数。

更具体地说，我需要知道哪个日期具有fld1和fld2的最独特组合。

如果table1有

1, 101, 101
2, 101, 101
3, 101, 101
4, 101, 102
5, 101, 102
6, 101, 103

表2有

1, 12/1/2010
2, 12/1/2010
3, 12/1/2010
1, 12/2/2010
2, 12/2/2010
4, 12/2/2010
5, 12/2/2010
6, 12/2/2010

我需要

12/1/2010, 1     'only 1 unique combinatin of fld1 and fld2
12/2/2010, 3     'only 3 unique combinations of fld1 and fld2

但我只需要“12/2/2010,3”输出，因为我只需要最大计数的日期和计数。

无法弄清楚如何在Commment中将其格式化为下面的正确答案 - 所以这里是MS Access 2003。

Select TOP 5 theDay, count(*) AS theCount 
FROM ( 
Select cdate(int(date_col)) As theDay 
From tbl1 Inner Join tbl2 on tbl1.id=tbl2.id 
Group By cdate(int(date_col)), fld1, fld2 
) As X 
Group By theDay 
Order By 2 Desc;

按日期返回TOP 5组合（忽略任何时间值）

Answer 1

在现代RDBMS中，您可以一次性计算不同的组合：

SELECT date_col, count(DISTINCT (fld1, fld2)) AS unique_combos
FROM   tbl1
JOIN   tbl2 USING (id)
GROUP  BY date_col;

如果不起作用，则分两步执行简单方法：

SELECT date_col, count(*) AS unique_combos
FROM  (
   SELECT date_col
   FROM   tbl1
   JOIN   tbl2 ON tbl1.id = tbl2.id
   GROUP  BY date_col, fld1, fld2
   ) AS x
GROUP  BY date_col;

第二个也应该在MS Access中工作。

最终@ user2184214的用途：

SELECT TOP 5 theDay, count(*) AS theCount
FROM  (
   SELECT cdate(int(date_col)) AS theDay
   FROM   tbl1 
   JOIN   tbl2 ON tbl1.id = tbl2.id
   GROUP  BY cdate(int(date_col)), fld1, fld2
   ) AS x
GROUP  BY theDay
ORDER  BY 2 DESC;

Answer 2

此查询为您提供所需内容：

SELECT
  date(dateAndTime) AS date,
  count(DISTINCT (fld1, fld2)) AS count
FROM table1
JOIN table2 ON table1.id = table2.id
GROUP BY 1
HAVING count(DISTINCT fld1, fld2) > 1
ORDER BY 2 DESC

计算两个表中的不同记录

2 个答案: