似乎无法使查询的结果唯一

Question

我的问题是，我只需要一次/唯一地对每种饮料（a.DRINK）做以下陈述，但我似乎做不到。

create table LOWEST as 
    select a.DRINK, min(b.PRICE), c.STREET, c.BLDG_NO
    from ALLDRINKS a
    left join SERVES b on a.DRINK = b.DRINK
    left join LOCATED c on b.PUB = c.PUB
    group by a.DRINK, c.STREET, c.BLDG_NO

我得到的当前结果

------------------------------------------------
| DRINK |min(b.PRICE)|     STREET    | BLDG_NO |
| VODKA |    7.10    | FAKE ST.      |   123   |
| VODKA |    4.50    | OAK Ave.      |    13   | 
| VODKA |    8.30    | Rail RD.      |    11   | 
| RUM   |    6.30    | Cherry RD.    |   131   | 
| RUM   |   10.30    | TEST Pl.      |    21   |  
------------------------------------------------

我期望的位置：

------------------------------------------------
| DRINK |min(b.PRICE)|     STREET    | BLDG_NO |
| VODKA |    4.50    | OAK Ave.      |    13   |  
| RUM   |    6.30    | Cherry RD.    |   131   |   
------------------------------------------------

如果您还可以解释需要做的事情背后的逻辑，那将对将来的工作大有帮助！先感谢您！-使用MYSQL Workbench 8.0

Answer 1

您可以使用表row_number()中的SERVES窗口函数来获取每个DRINK的最低价格，因此您不需要group by：

create table LOWEST as 
    select a.DRINK, b.PRICE, c.STREET, c.BLDG_NO
    from ALLDRINKS a
    left join (
      select t.* 
      from (
        select *, row_number() over (partition by DRINK order by price) rn
        from SERVES      
      ) t
      where t.rn = 1
    ) b on a.DRINK = b.DRINK
    left join LOCATED c on b.PUB = c.PUB

Answer 2

您想要serves中的行具有每个drink的最低价格，以及相应的location。如果是这样，您不应该考虑aggregation，而应该考虑filtering。

您可以使用相关子查询在left join中进行过滤：

create table lowest as 
select d.drink, s.price, c.street, l.bldg_no
from alldrinks d
left join serves s
    on  s.drink = d.drink
    and s.price = (select min(s1.price) from serves s1 where s1.drink = s.drink)
left join located l 
    on  l.pub = s.pub

请注意，我重命名了您的表别名，以使它们更有意义，这使查询更易于遵循。