Question

我有三个查询都返回1行和1列。我想将此作为1行和3列返回。

我的疑问

select count(distinct rt) from gpstable where rt >  GETDATE() - '1 day'::INTERVAL AND di like 'i';
select count(distinct rt) as count from gpstable where rt >  GETDATE() - '1 day'::INTERVAL group by di order by count limit 1;
select count(distinct rt) as count from gpstable where rt >  GETDATE() - '1 day'::INTERVAL group by di order by count DESC limit 1;

这就是我试过的

select userCount, min, max 
from 
(select count(distinct rt) as userCount from gpstablev2 where rt >  GETDATE() - '1 day'::INTERVAL AND di like 'UmTqUo1MQU8FgXfXXGNFh7vZnQN+bt2ThIQIDHNnmWU=') as userCount,
(select count(distinct rt) as min from gpstablev2 where rt >  GETDATE() - '1 day'::INTERVAL group by di order by count limit 1) as min,
(select count(distinct rt) as max from gpstablev2 where rt >  GETDATE() - '1 day'::INTERVAL group by di order by count DESC limit 1) as max;

我收到以下错误错误：gpstablev2中不存在列“count”[SQL State = 42703]

Answer 1

你很接近，你只需将个别陈述放入select列表，而不是from：

select (select count(distinct rt) from gpstable where rt >  GETDATE() - '1 day'::INTERVAL AND di = 'i') as user_count,
       (select count(distinct rt) from gpstable where rt >  GETDATE() - '1 day'::INTERVAL group by di order by count limit 1) as min_count,
       (select count(distinct rt) from gpstable where rt >  GETDATE() - '1 day'::INTERVAL group by di order by count DESC limit 1) as max_ount;

但我认为这可以简化并简化为表格上的单个查询：

select max(case when di = 'i' then di_count end) as user_count, 
       min(di_count) as min_count,
       max(di_count) as max_count
from (
    select di, 
           count(distinct rt) as di_count
    from gpstable 
    where rt >  current_date - interval '1' day
    group by di
)t ;

这只需要经过一次，而不是你尝试的三次。这样会更有效率。我只在一个非常小的数据集上测试过这个，所以可能是我错过了一些东西。

我使用ANSI SQL格式指定间隔interval '1' day，因为只要有等效版本，我就更喜欢标准语法（这也是我使用current_date代替getdate()的原因。 “一天”的持续时间实际上并不是必需的，因为默认单位是一天，因此current_date - 1也可以正常工作。

将三个查询合并为单个结果

1 个答案: