Question

我的数据跨越了多个月，我希望能够获得每天的平均值并将其分成适当的月份。例如，假设一个数据点是2/9/2010 - 3/8/2010，金额为1500.那么，查询应返回2010年2月的1071.4和3月的428.6。我希望有一个MySQL语句能够进行计算而不是我的PHP逻辑。感谢。

编辑（添加表定义）： start（datetime），end（datetime），use

编辑2：这是一些虚拟数据



DROP TABLE IF EXISTS `dummy_data`;
CREATE TABLE `dummy_data` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `start_date` date NOT NULL,
  `end_date` date NOT NULL,
  `data` double(15,4) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=MyISAM AUTO_INCREMENT=8 DEFAULT CHARSET=latin1;

-- ----------------------------
--  Records of `dummy_data`
-- ----------------------------
BEGIN;
INSERT INTO `dummy_data` VALUES ('1', '2010-01-01', '2010-02-02', '200.0000'), ('2', '2010-02-03', '2010-02-25', '250.0000'), ('3', '2010-02-26', '2010-03-08', '300.0000'), ('4', '2010-03-09', '2010-04-12', '210.0000'), ('5', '2010-04-13', '2010-05-10', '260.0000'), ('6', '2010-05-11', '2010-06-15', '310.0000'), ('7', '2010-06-16', '2010-07-20', '320.0000');
COMMIT;

Answer 1

您应该按MONTH（日期）功能选择SUM和分组，如下所示：

SELECT SUM(value), MONTH(date)
FROM TABLE
GROUP BY MONTH(date)

编辑：哎呀，我误解了这个问题，现在修改我的答案！

您需要使用一些更复杂的TSQL来获取月中的天数，找到平均值并设置到每个字段然后以月度格式显示

<强>更新创建一个函数，为您提供范围内的日期，例如我从Michael Baria创建的函数修改过的日期

CREATE FUNCTION [dbo].[GetDays](@StartDate DATETIME, @EndDate DATETIME)
RETURNS @MonthList TABLE(DayValue tinyint NOT NULL, MonthValue tinyint NOT NULL, YearValue int NOT NULL)
AS
BEGIN
--Variable used to hold each new date value
DECLARE @DateValue DATETIME

--Start with the starting date in the range
SET @DateValue=@StartDate

--Load output table with the month part of each new date
WHILE @DateValue <= @EndDate
BEGIN
    INSERT INTO @MonthList(DayValue, MonthValue,YearValue)
    SELECT DAY(@DateValue), MONTH(@DateValue), YEAR(@DateValue)

    --Move to the next day           
    SET @DateValue=DATEADD(dd,1,@DateValue)
END

--Return results
RETURN
END

GO

将您的桌子加入此功能，然后根据天数加总

SELECT SUM(data/DATEDIFF(dd,startDate,endDate)), M.MonthValue
FROM TABLE
JOIN (SELECT * FROM [dbo].[GetDays] (startDate,endDate)) M 

GROUP BY M.MonthValue

如果我得到一些样本数据，我可以稍微清理一下

Answer 2

此解决方案处理的[start_date, end_date]跨度只有一（1）天，最长可达十二（12）个月，但在十三（13）个月或更长时间内不正确：

CREATE TABLE integers (i INT NOT NULL);

INSERT INTO integers VALUES (0), (1), (2), (3), (4), (5), (6), (7), (8), (9);

CREATE VIEW hundreds AS
   SELECT iii.i * 100 + ii.i * 10 + i AS i
     FROM integers i JOIN integers ii JOIN integers iii;

-- We do not have CTEs, so we create a view
CREATE VIEW spans AS
   SELECT id, start_date, DATEDIFF(end_date, start_date) + 1 AS ndays, data
     FROM dummy_data;

   SELECT spans.id,
          month_name,
          spans.data * COUNT(month_name)/spans.ndays AS month_amount
     FROM spans
LEFT JOIN (SELECT id,
                  MONTH(start_date + INTERVAL i DAY) AS month_num,
                  MONTHNAME(start_date + INTERVAL i DAY) AS month_name
             FROM spans
             JOIN hundreds WHERE i < ndays) daybyday
       ON spans.id = daybyday.id
 GROUP BY id, month_name
 ORDER BY id, month_num;

输出如下：

+----+------------+---------------+
| id | month_name | month_amount  |
+----+------------+---------------+
|  1 | January    |  187.87878788 | 
|  1 | February   |   12.12121212 | 
|  2 | February   |  250.00000000 | 
|  3 | February   |   81.81818182 | 
...

我们使用DATEDIFF来确定源记录所代表的天数。然后，建立一个integers table，我们可以枚举特定范围内每一天的月份。从那里开始记录id和month_name的SQL聚合问题。

Answer 3

这确实将金额分开：

declare @start datetime set @start = '20100209'
declare @end datetime set @end = '20100308'
declare @avg float set @avg = 1500

select
datediff(day, @start, dateadd(day, 1-day(@end), @end)) * @avg / (datediff(day, @start, @end) + 1),
datediff(day, dateadd(day, -day(@end), @end), @end) * @avg / (datediff(day, @start, @end) + 1)

结果：

1071,42857142857    428,571428571429

虽然它变得有点复杂，因为你首先要检查日期是否实际上在不同月份，如果日期超过两个月你需要采用不同的方法。

如果跨越一个月以上的月度数据，如何计算MySQL的月平均值？

3 个答案: