似乎无法使查询结果唯一

Question

我的问题是我只需要在每杯饮料中做以下陈述 (a.DRINK) once/unique 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   |  
------------------------------------------------

我期待的地方：

------------------------------------------------
| 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完成的工作背后的逻辑，那将对以后非常有帮助！先感谢您！ -使用 MYSQL Workbench 8.0

Answer 1

可以在tableSERVES中使用row_number()window函数得到每个DRINK的最低价，所以不需要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 中具有每个 drink 最低价格的行，以及对应的 location。如果是这样你不应该考虑聚合，而是过滤.

您可以在 left join:

中使用相关子查询进行过滤

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 table 别名，因此它们更有意义，这使查询更容易理解。

似乎无法使查询结果唯一

Can't seem to make the results of query to be unique

mysql

sql

join

unique

greatest-n-per-group