Question

所以我有这张桌子

CREATE TABLE RaceResult(name char, time int, age int);
INSERT INTO RaceResult (name, time, age) VALUES
('Steve', '12', '33'),
('Tim', '34', '28'),
('Mark', '22', '37'),
('Tom', '21', '30'),
('Cliff', '13', '33'),
('Vini', '17', '28'),
('Matt', '10', '28'),
('Ben', '9', '29'),
('Brandon', '15', '14');

我想创建第4列，根据他们的年龄组将他们排在第1和第2位等等

年龄组

0 - 15岁
16 - 29岁
30岁以上

到目前为止我的选择声明是：

SELECT
CASE 
    WHEN age < '16' and  age > 0  
       THEN 'group1' 
    WHEN age < '30' and  age > 15  
       THEN 'group2' 
    WHEN age > '29'  
       THEN 'group3' 
       ELSE 0 
END as rank, *
FROM RaceResult
ORDER BY time

但不是排名列只是说'组1,2或3'我希望它输出如下：

            RESULTS
Name    Time    Age     Rank
=====   ====    ===     ====
Ben      9      29      1
Matt    10      28      2
Steve   12      33      1
Cliff   13      33      2
Brandon 15      14      1
Vini    17      28      3
Tom     21      30      3
Mark    22      37      4
Tim     34      28      4

真的很感激，如果有人能指出我正确的方向

Answer 1

试试这个。

;WITH cte
     AS (SELECT NAME,
                time,
                age,
                CASE
                  WHEN ( age / 15 ) + 1 < 3 THEN ( age / 15 ) + 1
                  ELSE 3
                END AS [group]
         FROM   RaceResult)
SELECT NAME,time,age,
       Row_number() OVER (partition BY [group] ORDER BY time ) [Rank]          
FROM   cte
ORDER  BY time

Answer 2

此查询应该执行您想要的操作：

SELECT
  age_group.*,
  @num := if (@age_group = age_group, @num + 1, 1) rank,
  @age_group := age_group as dummy
  FROM (
     SELECT
         RaceResult.*,
         @group := CASE 
                     WHEN age < '16' and  age > 0
                         THEN 'group1'
                     WHEN age < '30' and  age > 15
                         THEN 'group2'
                     WHEN age > '29'
                         THEN 'group3'
                     ELSE 0
                   END as age_group 
     FROM RaceResult,
     (select @age_group = '', @num = 1) init
     ORDER BY age_group, time
  ) age_group

解决方案基于以下链接的信息：

http://www.xaprb.com/blog/2006/12/02/how-to-number-rows-in-mysql/

SQL尝试根据年龄组超出年龄列制作排名列

2 个答案: