[英]How to select all rows that share a maximum value in a column using SQL
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 |
+----+---------+---------------------+
如何为每个用户选择具有的所有行:
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 |
+----+---------+---------------------+
我到目前为止所拥有的是这样的:
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();
但这无法按需工作。 我走的路好吗?
您走在正确的道路上。 在派生表中,您需要获取每个用户ID的最长日期,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上加入:
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();
_尝试这个:
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
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