Question

更新：重新格式化整个问题

我有一个包含个人F1比赛结果的数据库。每个驾驶员根据他们完成的位置获得积分。现在我想建立一个包含所有赛车的所有组合点的桌子。所以基本上是一个＆＃34;驱动程序排名＆＃34;表

我当然可以很容易地得到总结点，但是当多个驾驶员具有完全相同的总分时，我遇到的问题就出现了。在这些情况下，我需要检查哪一个驱动程序具有最佳（或更多）单个竞赛结果，因为他应该在结束表上方。

哦，在放置栏上＆＃34; 0＆＃34;意味着没有完成，这是最糟糕的结果。

简单示例：

Event  | Driver    | Placement | Points
--------------------------------------
Race 1 | Driver A  | 1         | 3
Race 1 | Driver B  | 2         | 2
Race 1 | Driver C  | 3         | 1
Race 2 | Driver A  | 3         | 3
Race 2 | Driver B  | 2         | 2
Race 2 | Driver C  | 1         | 1
Race 3 | Driver A  | 0         | 0
Race 3 | Driver B  | 0         | 0
Race 3 | Driver C  | 2         | 2

并要求完成表格

Driver    | Points
------------------
Driver C  | 4
Driver A  | 4
Driver B  | 4

订单是因为Driver C有一个第一名并且第二名完成。驾驶员A有一个第一名和第三名完成，其中驾驶员B只有第二名完成。（假设我的头很有效，那已经很晚了。）

简单示例http://sqlfiddle.com/#!9/16f9c5

期望的结果

| Driver        | Points |
|------------------------|
| N. Hulkenberg | 43     |
| F. Massa      | 43     |
| P. Wehrlein   |  5     |
| D. Kvyat      |  5     |
| M. Ericsson   |  0     |
| P. Gasly      |  0     |
| A. Giovinazzi |  0     |
| B. Hartley    |  0     |

从小提琴数据中我删除了除问题案例之外的所有内容。每个人都有至少一个相同的积分，并且领带破坏者从简单的最佳完成比赛到最佳结束次数（四个第六名胜利优于第二名胜利，如Hulkenberg和Massa的情况）。< / p>

结果表是因为：

1. N. Hulkenberg - has four 6th place finishes
2. F. Massa      - has two 6th place finishes
3. P. Wehrlein   - has 8th place finish
4. D. Kvyat      - has 9th place finish
5. M. Ericsson   - has 11th place finish
6. P. Gasly      - has 12th and 13th place finishes
7. A. Giovinazzi - has only one 12th place finish and nothing else
8. B. Hartley    - has 13th place finish

所以真正有问题的领带破坏者是[Hulkenberg vs Massa]和[Gasly vs Giovinazzi]

任何帮助表示感谢。

Answer 1

这应该可以做到没有太多模糊的伎俩。

SELECT     
    driver   
   ,SUM(points) AS points  
   ,CONCAT(GROUP_CONCAT(IF(placement = 0, 99, LPAD(placement,2,0)) ORDER BY IF(placement = 0, 99, placement) SEPARATOR ','), ',99') AS best 
FROM   
   f1 
GROUP BY   
   driver  
ORDER BY     
   points DESC,    
   best

或者，如果您想要隐藏展示位置排序数据列：

SELECT 
    driver
   ,SUM(points) AS points
FROM
    f1
GROUP BY
    driver
ORDER BY 
    points DESC
   ,CONCAT(GROUP_CONCAT(IF(placement = 0, 99, LPAD(placement,2,0)) ORDER BY IF(placement = 0, 99, placement) SEPARATOR ','), ',99')

它只是利用字母数字排序来确定在点数相关的情况下的最佳位置。

Answer 2

尝试自我加入。

select t1.driver, points, min_placement 
from (select driver, sum(points) points from f1 group by driver) t1 inner join
     (select driver, min(placement) min_placement from f1 group by driver) t2 
  on t1.driver = t2.driver
order by points desc, min_placement asc

在您的测试数据中，大多数司机的展示位置最低为零，因此您无法看到您正在寻找的结果。

Answer 3

我认为你在最高成绩后至少需要第二次抢七：

SELECT
    driver,
    SUM(points) as points,
    CASE WHEN MAX(placement) = 0 THEN 99 ELSE MAX(placement) END AS highest_finish,
    -- or maybe this would work?
    SUM(CASE WHEN placement <> 0
        THEN POWER(20, placement) ELSE 0 END) AS weighted_finish 
FROM f1
GROUP BY driver
ORDER BY points DESC, highest_finish

你可能也会幸运得到这样的表达式：

SELECT f1.driver,
    MAX((100 - f1.placement) * (
        SELECT count(*) FROM f1 f2
        WHERE f2.driver = f1.driver 
          AND f2.placement = f1.placement
          AND f2.placement > 0
        )) AS highest_placement_weight
FROM f1
GROUP BY f1.driver

当总计总值相等时，MYSQL分类

3 个答案: