Question

表格结构：

注册：

uuid | name | total

评分：

uuid | type | rate

Registration_Rate：

registration | rate

初始请求是：

select * from registration r
join registration_rate rr on rr.registration = r.uuid
join rate rt on rt.uuid = rr.rate
group by r.name,  rt.type

来自两个表（注册和费率）的我的SQL结果是：

  uuid | name | rate | type
     1 | AAA  | 15   | U
     2 | BBB  | 20   | U
     3 | CCC  | 300  | F
     4 | AAA  | 250  | F

我想有这样的东西（如果存在费率类型'F'则显示相反）

  uuid | name | rate | type
     2 | BBB  | 20   | U
     3 | CCC  | 300  | F
     4 | AAA  | 250  | F

由于

已编辑：

我尝试了另一种有效的解决方案

select uuid, name, rate, (case rt.type when 2 then 2 else 1 end ) as type
from registration r
join registration_rate rr on rr.registration = r.uuid
join rate rt on rt.uuid = rr.rate
group by r.name,  rt.type

Answer 1

如果是F行则返回它。或者，使用NOT EXISTS验证没有其他具有相同名称的行具有F。

select t1.*
from tablename t1
where type = 'F'
   or not exists (select * from tablename t2
                  where t2.name = t1.name
                    and t2.type = 'F')

替代解决方案：

select t1.*
from tablename t1
    join (select name, min(type) type
          from tablename
          group by name) t2
    ON t1.name = t2.name and t1.type = t2.type

Answer 2

试试这个（我建议主要想法）

 SELECT t.uuid, 
       t.name, 
       IFNULL(MAX(t.F_type), MAX(t.not_F_type)) AS "type", 
       IFNULL(MAX(t.F_rate), MAX(t.not_F_rate)) AS "rate"
FROM
(
    SELECT r.uuid, 
           r.name,
           CASE rt.type WHEN 'F' THEN rt.type END AS F_type, 
           CASE WHEN rt.type <> 'F' THEN rt.type END AS not_F_type,
           CASE rt.type WHEN 'F' THEN rt.rate END AS F_rate, 
           CASE WHEN rt.type <> 'F' THEN rt.rate END AS not_F_rate
    FROM registration AS r
    JOIN registration_rate AS rr ON rr.registration = r.uuid
    JOIN rate AS rt ON rt.uuid = rr.rate
) as t
GROUP BY t.uuid, t.name;

因此，您需要根据您的规则（如果比率的类型'F'存在，然后显示而不是其他）将适当的列（“rate”，“type”）拆分为使用case语句的两个新的单独列：第一个column包含F类型的值，第二列包含其他类型的值。我为“type”和“rate”列做了这个。然后我使用group by，aggregation functions和IFNULL语句将这些列（和记录）粘在一起（你可以在这里使用其他语句：case，IF等）。

据我所知，这就是你所需要的。

按优先级选择

2 个答案: