6

我读过这篇文章：http ://www.xaprb.com/blog/2006/12/07/how-to-select-the-firstleastmax-row-per-group-in-sql/并搜索其他问题

我有一张类似的桌子：

| table.id    | USER.id
----------------------------------------------
| 1           | 101
| 2           | 101
| 3           | 101
| 4           | 101 
| 5           | 101
| 6           | 101
| 7           | 101 
| 8           | 101
| 9           | 101 
| 10          | 101
| 11          | 102
| 12          | 102 
| 13          | 102
| 14          | 102
| 15          | 103 
| 16          | 103
| 17          | 103
| 18          | 103
| 19          | 103
| 20          | 103
| 21          | 103
| 22          | 103 
| 23          | 103
| 24          | 104
| 25          | 104 
| 26          | 104
| 27          | 104
| 28          | 104
| 29          | 104
| 30          | 105
| 31          | 105
| 32          | 105 
| 33          | 106
| 34          | 106

我正在尝试获取按 user.id 分组的 table.id 的计数，如果 user.id 的计数超过 7，则仅将结果显示为 7（也将计数结果限制为 7）。

在此示例中，结果应为：

| USER.id         | count of table.ID
----------------------------------------
| 101             | 7
| 102             | 4
| 103             | 7
| 104             | 6
| 105             | 3
| 106             | 2

我试过了：

SELECT USERid, COUNT(table.id)
FROM table
WHERE table.id IN (select top 7 table.id from table)
GROUP BY USERid

和

SELECT USERid, COUNT(table.id)
FROM table
WHERE (
    SELECT COUNT(table.ID) FROM table as t
    WHERE t.id = t.id AND t.USERid <= table.USERid
    ) <= 7
GROUP BY USERid
4

3 回答 3

7

您可以简化查询，并使用LEAST函数

SELECT USERid, LEAST(7, COUNT(*))
FROM table
GROUP BY USERid

从您评论中的问题

SELECT SUM(countByUser) 
FROM
 (SELECT LEAST(7, COUNT(*)) as countByUser
  FROM table
  GROUP BY USERid) c

SqlFiddle

于 2013-10-03T20:42:12.420 回答
1 
SELECT userid,
CASE
  WHEN COUNT(*) > 7 THEN 7
  ELSE COUNT(*)
END AS Qty
FROM tbl
GROUP BY userid
于 2013-10-03T20:45:42.057 回答
0 
SELECT USERid, case when COUNT(table.id) > 7 
                    then 7 
                    else COUNT(table.id) 
               end
FROM table
GROUP BY USERid
于 2013-10-03T20:42:00.810 回答