Question

我具有以下结构http://sqlfiddle.com/#!9/f8341c 从这些记录中，我的预期输出如下：

Name|MinCost |MaxCost | Open | Close | Date
ABC |13.6    | 15.3   | 14.1 | 14.2  | 2015-12-02
DEF |93.2    | 96.3   | 93.7 | 95.4  | 2015-12-02 
ABC |15.1    | 15.6   | 15.1 | 15.2  | 2015-12-03
DEF |97.2    | 97.7   | 97.7 | 97.7  | 2015-12-03

现在我已经完成了

SELECT t1.name as 'Name',t1.min as 'MinCost',t1.max as 'MaxCost',t2.open ,t3.close,t1.times as 'Date'
  FROM(
        select name, max(cost) as 'max',min(cost) as 'min', date(time) as 'times'  
        from providers 
        group by  times,name 
        order by times
   ) AS t1
   JOIN (
     select name,cost as 'open',  time  
     from providers 
     where TIME(time) = '00:00:00'   
     group by time,name
     order by time
   ) as t2 on t1.name=t2.name
   JOIN (
       select name,cost as 'close', time  
       from providers 
       where TIME(time) = '23:59:59'   
       group by time,name
       order by time
   ) as t3 on t2.name=t3.name 
   GROUP BY t1.times,t1.name
   ORDER BY t1.times,t1.name

这给了我以下输出

| Name | MinCost | MaxCost | open | close |       Date |
|------|---------|---------|------|-------|------------|
|  ABC |    13.6 |    15.3 | 14.1 |  14.2 | 2015-12-02 |
|  DEF |    93.2 |    96.3 | 97.7 |  95.4 | 2015-12-02 |
|  ABC |    15.1 |    15.6 | 14.1 |  14.2 | 2015-12-03 |
|  DEF |    97.2 |    97.7 | 97.7 |  95.4 | 2015-12-03 |

我需要更正查询什么？

Answer 1

由于您的姓名不是唯一的，因此需要在联接time子句中添加on，以便使用name和time获得唯一的值。

下面的查询应该可以正常工作，Demo on your fiddle以获取预期的输出

  SELECT t1.name as 'Name',t1.min as 'MinCost',t1.max as 'MaxCost',t2.open ,t3.close,t1.times as 'Date'
  FROM(
        select name, max(cost) as 'max',min(cost) as 'min', date(time) as 'times'  
        from providers 
        group by  times,name 
        order by times
   ) AS t1
   JOIN (
     select name,cost as 'open',  date(time) as 'times' 
     from providers 
     where TIME(time) = '00:00:00'   
     group by time,name
     order by time
   ) as t2 on t1.name=t2.name and t1.times = t2.times
   JOIN (
       select name,cost as 'close', date(time) as 'times' 
       from providers 
       where TIME(time) = '23:59:59'   
       group by time,name
       order by time
   ) as t3 on t2.name=t3.name  and t2.times = t3.times
   GROUP BY t1.times,t1.name
   ORDER BY t1.times,t1.name

Answer 2

您在这里：

select
  d.name, d.min_cost, d.max_cost,
  o.cost as open,
  c.cost as close,
  d.d as date
from (
  select
    name,
    min(cost) as min_cost,
    max(cost) as max_cost,
    min(time) as open_time,
    max(time) as close_time,
    date(time) as d
  from providers
  group by name, date(time)
) d
join providers o on o.name = d.name and o.time = d.open_time
join providers c on c.name = d.name and c.time = d.close_time
order by d.d, d.name

结果：

name  min_cost  max_cost  open  close  date
----  --------  --------  ----  -----  ----------
ABC       13.6      15.3  14.1   14.2  2015-12-02
DEF       93.2      96.3  93.7   95.4  2015-12-02
ABC       15.1      15.6  15.1   15.2  2015-12-03
DEF       97.2      97.7  97.7   97.4  2015-12-03

请注意，由于开盘价为97.7，因此最后一行的预期结果有些错误。

如何获得正确的最小和最大值

2 个答案: