Question

我需要为10天的间隔计算单个用户的访问次数。

我总是看到上次访问的日期。

我已经尝试过这个查询，但它有问题。

mysql> SELECT
    DATE_FORMAT(MIN(`theDate`), '%d/%m/%Y') AS 'First Access',
    DATE_FORMAT(MAX(`theDate`), '%d/%m/%Y') AS 'Last Access',
    COUNT(*) AS 'Number',
        TheUser
FROM
    `doTable`
WHERE
    `theDate` BETWEEN MAKEDATE(YEAR(CURDATE()),1)
AND DATE_SUB(MIN(`theDate`), INTERVAL 10 DAY)
GROUP BY
    TheUser
HAVING
    COUNT(*) > 1
ORDER BY
    COUNT(*) DESC;
1111 - Invalid use of group function

如果我使用日期值，则查询有效。

mysql> SELECT
    DATE_FORMAT(MIN(`theDate`), '%d/%m/%Y') AS 'First Access',
    DATE_FORMAT(MAX(`theDate`), '%d/%m/%Y') AS 'Last Access',
    COUNT(*) AS 'Number',
        TheUser
FROM
    `doTable`
WHERE
    `theDate` BETWEEN '2015-01-01'
AND DATE_SUB('2015-01-22', INTERVAL 10 DAY)
GROUP BY
    TheUser
HAVING
    COUNT(*) > 1
ORDER BY
    COUNT(*) DESC;

输出：

+---------+------------+------------+--------+
| TheUser | First      | Last       | Number |
+---------+------------+------------+--------+
| s074    | 05/01/2015 | 05/01/2015 |     18 |
| x066    | 02/01/2015 | 12/01/2015 |     17 |
| f918    | 03/01/2015 | 03/01/2015 |     17 |
| l371    | 01/01/2015 | 02/01/2015 |     15 |
| b779    | 01/01/2015 | 02/01/2015 |     14 |
+---------+------------+------------+--------+
5 rows in set

编辑＃1

我已将条件移动到having子句，但输出为空。

mysql> SELECT
    DATE_FORMAT(MIN(`theDate`), '%d/%m/%Y') AS 'First Access',
    DATE_FORMAT(MAX(`theDate`), '%d/%m/%Y') AS 'Last Access',
    COUNT(*) AS 'Number',
    TheUser,
    `DATA`
FROM
    `doTable`       
GROUP BY
    TheUser
HAVING
    'Number' > 1
AND `theDate` BETWEEN MAKEDATE(YEAR(CURDATE()),1)
AND DATE_SUB(MIN(`DATA`), INTERVAL 10 DAY)
ORDER BY
    'Number' DESC;

MySQL 1111 - 使用最小日期时无效使用组功能

0 个答案: