Question

这是我的sql fiddle http://sqlfiddle.com/#!2/9e6691/3

[Quater 1:01] [Quater 2:04] [Quater 3:07] [Quater 4：10]

我正在开发Google图表，我正在为单独的项目创建图表。图表由两行组成。一行将提到该项目的ratepersquarefeet。其他线路希望得到关于四分之一和公共位置的平均值。

编码如下：

CREATE TABLE IF NOT EXISTS `ratepersqft` (
  `price_id` int(11) NOT NULL auto_increment,
  `project_id` int(11) default NULL,
  `date` varchar(255) default NULL,
  `rate_per_sqft` double default NULL,
  `common_location` varchar(255) default NULL,
  PRIMARY KEY  (`price_id`));

INSERT INTO `ratepersqft` (`price_id`, `project_id`, `date`, `rate_per_sqft`, `common_location`) VALUES
    (1, 26, '2012-01-17 ', 2800, 'ECR'),
    (2, 54, '2012-04-17 ', 1550, 'ECR'),
    (3, 53, '2012-07-17 ', 1850, 'ECR'),
    (4, 58, '2012-10-17 ', 4425, 'ECR'),
    (5, 85, '2012-01-17 ', 0, 'ECR'),
    (6, 114, '2012-04-17 ', 2650, 'ECR'),
    (7, 131, '2012-07-17 ', 2400, 'ECR'),
    (8, 134, '2012-10-17 ', 2900, 'ECR'),
    (9, 140, '2012-01-17 ', 2500, 'ECR'),
    (10, 368, '2012-04-17 ', 1650, 'ECR'),
    (11, 369, '2012-07-17 ', 1800, 'ECR'),
    (12, 370, '2012-10-17 ', 2000, 'ECR'),
    (13, 26, '2012-04-17 ', 2800, 'ECR'),
    (14, 26, '2012-07-17 ', 2800, 'ECR'),
    (15, 26, '2012-10-17 ', 2800, 'ECR');

例如。在小提琴中，我将项目ID 26作为其在ECR中的共同位置

图表中的第一行将与该项目中的ratepersquarefeet有关。还有一些其他项目将ECR作为共同位置。我需要计算这些项目相对于上述日期的平均值。

对于Eg。 price_id：1,5,9每平方英尺的费率是2800,0,2500 First Quater的平均值是1766.就像我需要为每个季度计算一样。

Answer 1

SELECT ROUND(avg(rate_per_sqft)) as RatePerSqFt 
  FROM ratepersqft
 WHERE common_location = 'ECR' AND
       1 <= MONTH(date) AND MONTH(date) <3  
       group by common_location
UNION
SELECT ROUND(avg(rate_per_sqft)) as RatePerSqFt 
  FROM ratepersqft
 WHERE common_location = 'ECR' AND
       3 <= MONTH(date) AND MONTH(date)<6  
       group by common_location
UNION
SELECT ROUND(avg(rate_per_sqft)) as RatePerSqFt 
  FROM ratepersqft
 WHERE common_location = 'ECR' AND
       6 <= MONTH(date) AND MONTH(date)<9  
       group by common_location
UNION
SELECT ROUND(avg(rate_per_sqft)) as RatePerSqFt 
  FROM ratepersqft
 WHERE common_location = 'ECR' AND
       9 <= MONTH(date) AND MONTH(date)<=12
       group by common_location;

Answer 2

接受的答案不是一种令人费解的说法：

SELECT QUARTER(date) dt
     , common_location
     , ROUND(avg(rate_per_sqft)) as RatePerSqFt 
  FROM ratepersqft
 GROUP 
    BY dt
     , common_location  ;

Answer 3

您可以通过在子选择中按季度/位置计算平均值然后再次加入表格来实现此目的。

这是我提出的查询：

select * 
from ratepersqft r
join (
    SELECT 
      quarter(`date`) as quarter, 
      year(`date`) as year, 
      avg(rate_per_sqft) as QuarterAverage,
      common_location
    FROM ratepersqft
    group by 
      quarter(`date`), 
      year(`date`),
      common_location
  ) averages
  on year(r.`date`) = averages.year
  and quarter(r.`date`) = averages.quarter
  and r.common_location = averages.common_location

WHERE project_id = 26

SQL小提琴：http://sqlfiddle.com/#!2/9e6691/23

获取与SQL中的日期相关的平均值

3 个答案: