Question

我想查询最高点的排名，如果点相同，检查完成的时间是最短的。我的桌子是

 +-------------------------------------------------+
 | | id | user_id | point | finishtime  | week  | |
 +-------------------------------------------------+
 | | 1  | G1      | 1560  |    55       |    1  | |
 | | 2  | G1      | 1560  |    43       |    1  | |
 | | 3  | G1      | 1530  |    55       |    1  | |
 | | 4  | G2      | 1760  |    45       |    1  | |
 | | 5  | G3      | 1760  |    46       |    1  | |
 | | 6  | G3      | 1330  |    25       |    2  | |
 | | 7  | G4      | 360   |    65       |    1  | |
 | | 8  | G2      | 1760  |    50       |    1  | |

预期结果是

 +-------------------------------------------------+
 | | id | user_id | point | finishtime  | week  | |
 +-------------------------------------------------+
 | | 4  | G2      | 1760  |    45       |    1  | |
 | | 5  | G3      | 1760  |    46       |    1  | |
 | | 2  | G1      | 1560  |    43       |    1  | |
 | | 7  | G4      | 360   |    65       |    1  | |
 | | 6  | G3      | 1330  |    25       |    2  | |

我尝试选择max作为该点，但它不会花费最短的完成时间。

我需要唯一的user_id结果组，按最高点获得订单，最短的完成时间。

我需要使用if else语句吗？

Answer 1

这是一个可能的答案：

SELECT t1.* FROM t t1
JOIN (
  SELECT t1.user_id, t1.week, t1.point, min(t1.finishtime) ft FROM t t1
  LEFT JOIN t t2
  ON t1.week = t2.week AND t1.user_id = t2.user_id AND t1.point < t2.point
  WHERE t2.point IS NULL
  GROUP BY t1.user_id, t1.week, t1.point
) t2 ON t1.user_id = t2.user_id AND t1.week = t2.week AND
        t1.point = t2.point AND t1.finishtime = t2.ft
ORDER BY t1.week, t1.point DESC, t1.finishtime

<强>结果：

| ID | USER_ID | POINT | FINISHTIME | WEEK |
|----|---------|-------|------------|------|
|  4 |      G2 |  1760 |         45 |    1 |
|  5 |      G3 |  1760 |         46 |    1 |
|  2 |      G1 |  1560 |         43 |    1 |
|  7 |      G4 |   360 |         65 |    1 |
|  6 |      G3 |  1330 |         25 |    2 |

小提琴here。

这基本上是每组最大的双重问题，因为你需要首先获得最高点的那些，然后获得最小完成时间的那些。

另一种解决方案是通过...来获取双组，但这将涉及第三个嵌套级别并试图避免它，所以选择了左连接解决方案。

Answer 2

以下是使用JOINS

进行此操作的方法

 SELECT
    DISTINCT
    t1.user_id,
    max(t2.point),
    min(t2.finishtime),
    min(t1.week)
 FROM
    t t1
    JOIN t t2 ON ( t1.user_id = t2.user_id )
 GROUP BY 
    t1.user_id
 HAVING
    MAX( t1.point ) = MAX( t2.point )
    AND MIN( t1.finishtime ) = MIN( t2.finishtime )
 ORDER BY 
    t1.user_id

结果：

| USER_ID | MAX(T2.POINT) | MIN(T2.FINISHTIME) | MIN(T1.WEEK) |
|---------|---------------|--------------------|--------------|
|      G1 |          1560 |                 43 |            1 |
|      G2 |          1760 |                 45 |            1 |
|      G3 |          1760 |                 25 |            1 |
|      G4 |           360 |                 65 |            1 |

感谢Mosty的SQLFIDDLE

Answer 3

SELECT id, user_id, point, finishtime, week
FROM (  SELECT id, user_id, point, finishtime, week
        FROM table_name
        ORDER BY point DESC, finishtime ASC) AS h
GROUP BY user_id, week
ORDER BY week ASC

这可能是使用未记录的行为执行此操作的最短方法之一。请阅读以下内容。

按最高点和最短时间分组

3 个答案: