Mysql 如何使用SQL选择列中共享最大值的所有行

我有一张这样的桌子：

+----+---------+---------------------+
| id | user_id |     start_date      |
+----+---------+---------------------+
|  1 |       1 | 2014-02-01 00:00:00 |
|  2 |       1 | 2014-01-01 00:00:00 |
|  3 |       2 | 2014-01-01 00:00:00 |
|  4 |       2 | 2014-01-01 00:00:00 |
|  5 |       3 | 2015-01-01 00:00:00 |
+----+---------+---------------------+

SELECT t.* FROM ticket t
    JOIN (
        SELECT start_date, MAX(start_date) FROM ticket /* GROUP BY user_id */
    ) highest
    ON t.start_date = highest.start_date
    WHERE t.start_date <= NOW();

如何为每个用户选择具有以下内容的所有行：

开始日期在现在和之前 最长开始日期 因此，对于示例行，输出应为：

+----+---------+---------------------+
| id | user_id |     start_date      |
+----+---------+---------------------+
|  1 |       1 | 2014-02-01 00:00:00 | // this is a single maximum date within that user
|  3 |       2 | 2014-01-01 00:00:00 | // these two share maximum start date
|  4 |       2 | 2014-01-01 00:00:00 |
+----+---------+---------------------+

到目前为止，我得到的是这样的东西：

+----+---------+---------------------+
| id | user_id |     start_date      |
+----+---------+---------------------+
|  1 |       1 | 2014-02-01 00:00:00 |
|  2 |       1 | 2014-01-01 00:00:00 |
|  3 |       2 | 2014-01-01 00:00:00 |
|  4 |       2 | 2014-01-01 00:00:00 |
|  5 |       3 | 2015-01-01 00:00:00 |
+----+---------+---------------------+

SELECT t.* FROM ticket t
    JOIN (
        SELECT start_date, MAX(start_date) FROM ticket /* GROUP BY user_id */
    ) highest
    ON t.start_date = highest.start_date
    WHERE t.start_date <= NOW();

---

但这并没有达到预期效果。我走的路好吗

你的思路是对的。 在派生表中，需要获取每个用户id的最大日期，因此：

SELECT user_id, 
      MAX(start_date) as MaxDate
      FROM ticket 
      GROUP BY user_id

然后，您可以在开始日期和用户id加入：

SELECT t.* FROM ticket t
    JOIN (
      SELECT user_id, 
      MAX(start_date) as MaxDate
      FROM ticket 
      GROUP BY user_id
    ) highest
    ON t.start_date = highest.maxdate
  and t.user_id = highest.user_id
    WHERE t.start_date <= NOW();

---

在某种程度上，你走对了方向。 在派生表中，需要获取每个用户id的最大日期，因此：

SELECT user_id, 
      MAX(start_date) as MaxDate
      FROM ticket 
      GROUP BY user_id

然后，您可以在开始日期和用户id加入：

SELECT t.* FROM ticket t
    JOIN (
      SELECT user_id, 
      MAX(start_date) as MaxDate
      FROM ticket 
      GROUP BY user_id
    ) highest
    ON t.start_date = highest.maxdate
  and t.user_id = highest.user_id
    WHERE t.start_date <= NOW();

_试试这个：

SELECT T.* FROM ticket AS T
JOIN (SELECT
     [User_Id]
    ,MAX([Start_Date])  AS Start_Date
FROM ticket
WHERE Start_Date <= GETDATE()
GROUP BY User_Id) AS Grouped    ON T.User_Id = Grouped.User_Id AND T.Start_Date = Grouped.Start_Date
ORDER BY Id
DROP TABLE #This

_试试这个：

SELECT T.* FROM ticket AS T
JOIN (SELECT
     [User_Id]
    ,MAX([Start_Date])  AS Start_Date
FROM ticket
WHERE Start_Date <= GETDATE()
GROUP BY User_Id) AS Grouped    ON T.User_Id = Grouped.User_Id AND T.Start_Date = Grouped.Start_Date
ORDER BY Id
DROP TABLE #This