Question

我有2张桌子。 Person表和activity_log表

每个人都有很多活动。

我想“选择”显示其上一个活动的activity_name和activity_date的“人员”列表。（并非所有活动）

人员表有person_id，姓名，电子邮件

活动表包含person_id，activity_id，activity_date，activity_name

提前感谢您的帮助。

Answer 1

假设您的ID是自动编号的，您可以这样做：

SELECT pe.person_id,
       pe.name,
       al.activity_name,
       al.activity_date
FROM   person pe
       LEFT JOIN (SELECT p.person_id,
                         Max(a.activity_id) activity_id
                  FROM   person p
                         LEFT JOIN activity_log a
                           ON ( p.person_id = a.person_id )
                  GROUP  BY p.person_id) AS LAST
         ON pe.person_id = LAST.person_id
       LEFT JOIN activity_log al
         ON LAST.activity_id = al.activity_id

但是，用户可能会比较新的活动更晚地输入过去的活动，然后这会失败，你必须这样：

SELECT LAST.person_id,
       LAST.name,
       LAST.activity_date,
       (SELECT activity_name
        FROM   activity_log al
        WHERE  al.person_id = LAST.person_id
               AND al.activity_date = LAST.activity_date) activity_name
FROM   (SELECT p.person_id,
               Max(p.name)          AS name,
               Max(a.activity_date) activity_date
        FROM   person p
               LEFT JOIN activity_log a
                 ON ( p.person_id = a.person_id )
        GROUP  BY p.person_id) AS LAST

但是这仍然存在问题：由于MySQL在子查询中不允许LIMIT，如果同一个人有两个活动具有相同的activity_date并且这是最新日期，则查询将失败。

Answer 2

子查询应该可以解决问题：

SELECT p.person_id, p.name, p.email, a.activity_id, a.activity_date, a.activity_name
FROM Person p
LEFT JOIN Activity a
ON p.person_id = a.person_id
WHERE a.activity_date = (SELECT MAX(a1.activity_date)
               FROM activity a1
               WHERE a.person_id = a1.person_id
               GROUP BY person_id) OR activity_date IS NULL;

mysql表连接聚合

2 个答案: