MySQL选择具有最小值的所有子查询行

Question

我想选择给定字段中具有最小值的子查询的所有行。以下是我迄今为止尝试过的一些技术示例：

-- 1.
select
    id, min(foo)
from
    (select 1 AS id, 2 AS foo
    union select 2 AS id, 2 AS foo
    union select 3 AS id, 3 AS foo) a;

-- 2.
select
    min(foo)
from
    (     
          select 1 AS id, 2 AS foo, 0 AS const
    union select 2 AS id, 2 AS foo, 0 AS const
    union select 3 AS id, 3 AS foo, 0 AS const) a
group by const;

-- 3.
select
    id
from
    (select 1 AS id, 2 AS foo
    union select 2 AS id, 2 AS foo
    union select 3 AS id, 3 AS foo) a
where id = (select id from a where min(foo) = foo);

-- 4.
select
    id
from
    (select 1 AS id, 2 AS foo
    union select 2 AS id, 2 AS foo
    union select 3 AS id, 3 AS foo) a
where foo = (select min(foo));

-- 5.
select r.*
from
(
    select min(foo) t
    from
        (select 1 AS id, 2 AS foo
        union select 2 AS id, 2 AS foo
        union select 3 AS id, 3 AS foo) a
) m
INNER JOIN a ON m.t = r.foo;

我正在处理的实际查询与示例类似，因为它由几个较小的查询联合在一起组成。这里的总体目标是根据关联表 k 的字段在 central 表中查找一行，其中 k 是最高优先级表。结果是来自相似（但不同表）的行的一种树状视图。

我已经提到了这一点，以防万一有人看到我正在绕道而行，他们可以对大局有所了解。但现在我的角度是通过在子查询中的字段上取最小值来选择。

Answer 1

使用order by 和limit：

select t.*
from t
order by foo
limit 1;

注意：即使有重复，这也只会返回具有最小值的一行。 t 是您的子查询或表。

如果你想要所有这些，那么你需要包含两次表定义：

select t.*
from t
where t.foo = (select min(t2.foo) from t t2);