选择最小列值并仅查询不同的相应行

Question

我试图用多个非唯一外键来查询一个表。我想选择“存储”与最小距离（英里），所以可能有多个商店的距离不同，我想有一个明显的一行最低英里

SELECT 
    m.store1, m.miles, mitable.region, mitable.terminal, mitable.city, mitable.state 
FROM 
    (SELECT 
     store1, MIN (miles) as miles 
    FROM 
     mitable 
    GROUP BY 
     region, store1) m 
INNER JOIN 
    mitable ON m.store1 = mitable.store1 
      AND m.miles = mitable.miles 

眼下，这是拉动了错误的数据，我...我卡住了！

Answer 1

只需使用row_number()：

select t.* 
from (select t.*, 
      row_number() over (partition by store1 order by miles asc) as seqnum 
     from mitable t 
    ) t 
where seqnum = 1; 

编辑：

如果你想用里程最少的行，那么你可以做：

select top 1 t.* 
from mitable t 
order by miles asc; 

这正好返回一行。如果你想要绑定多行，则使用top with ties。

Answer 2

呦很接近。只要删除region在group by

SELECT 
    m.store1, m.miles, mitable.region, mitable.terminal, mitable.city, mitable.state 
FROM 
    (SELECT 
     store1, MIN (miles) as miles 
    FROM 
     mitable 
    GROUP BY 
     store1 
     --region 
    ) m 
INNER JOIN 
    mitable ON m.store1 = mitable.store1 
      AND m.miles = mitable.miles