当前位置: 代码迷 >> Sql Server >> 全天各个时间段产品销量状况统计
  详细解决方案

全天各个时间段产品销量状况统计

热度:69   发布时间:2016-04-24 08:57:13.0
全天各个时间段产品销量情况统计

数据库环境:SQL SERVER 2005

现有一个产品销售实时表,表数据如下:

字段name是产品名称,字段type是销售类型,1表示售出,2表示退货,字段num是数量,字段ctime是操作时间。

要求:

  在一行中统计24小时内所有货物的销售(售出,退货)数据,把日期考虑在内。

分析:

  这实际上是行转列的一个应用,在进行行转列之前,需要补全24小时的所有数据。补全数据可以通过系统的数字辅助表

spt_values来实现,进行行转列时,根据type和处理后的ctime分组即可。

1.建表,导入数据

CREATE TABLE snake (name VARCHAR(10 ),type INT,num INT, ctime DATETIME )INSERT INTO snake VALUES(' 方便面', 1,10 ,'2015-08-10 16:20:05')INSERT INTO snake VALUES(' 香烟A ', 2,2 ,'2015-08-10 18:21:10')INSERT INTO snake VALUES(' 香烟A ', 1,5 ,'2015-08-10 20:21:10')INSERT INTO snake VALUES(' 香烟B', 1,6 ,'2015-08-10 20:21:10')INSERT INTO snake VALUES(' 香烟B', 2,9 ,'2015-08-10 20:21:10')INSERT INTO snake VALUES(' 香烟C', 2,9 ,'2015-08-10 20:21:10')
View Code

2.补全24小时的数据

/*枚举0-23自然数列*/WITH    x0          AS ( SELECT   number AS h               FROM     master..spt_values               WHERE    type = 'P'                        AND number >= 0                        AND number <= 23             ),/*找出表所有的日期*/        x1          AS ( SELECT DISTINCT                        CONVERT(VARCHAR(100), ctime, 23) AS d               FROM     snake             ),/*补全所有日期的24小时*/        x2          AS ( SELECT   x1.d ,                        x0.h               FROM     x1                        CROSS JOIN x0             ),        x3          AS ( SELECT   name ,                        type ,                        num ,                        DATEPART(hour, ctime) AS h               FROM     snake             ),/*整理行转列需要用到的数据*/        x4          AS ( SELECT   x2.d ,                        x2.h ,                        x3.name ,                        x3.type ,                        x3.num               FROM     x2                        LEFT JOIN x3 ON x3.h = x2.h             )
View Code

3.行转列

SELECT  ISNULL([0], 0) AS [00] ,            ISNULL([1], 0) AS [01] ,            ISNULL([2], 0) AS [02] ,            ISNULL([3], 0) AS [03] ,            ISNULL([4], 0) AS [04] ,            ISNULL([5], 0) AS [05] ,            ISNULL([6], 0) AS [06] ,            ISNULL([3], 7) AS [07] ,            ISNULL([8], 0) AS [08] ,            ISNULL([9], 0) AS [09] ,            ISNULL([10], 0) AS [10] ,            ISNULL([3], 11) AS [11] ,            ISNULL([12], 0) AS [12] ,            ISNULL([13], 0) AS [13] ,            ISNULL([14], 0) AS [14] ,            ISNULL([3], 15) AS [15] ,            ISNULL([16], 0) AS [16] ,            ISNULL([17], 0) AS [17] ,            ISNULL([18], 0) AS [18] ,            ISNULL([19], 15) AS [19] ,            ISNULL([20], 0) AS [20] ,            ISNULL([21], 0) AS [21] ,            ISNULL([22], 0) AS [22] ,            ISNULL([23], 15) AS [23] ,            type ,            d AS date    FROM    ( SELECT    d ,                        h ,                        type ,                        num              FROM      x4            ) t PIVOT( SUM(num) FOR h IN ( [0], [1], [2], [3], [4], [5], [6],                                           [7], [8], [9], [10], [11], [12],                                           [13], [14], [15], [16], [17], [18],                                           [19], [20], [21], [22], [23] ) ) t    WHERE   type IS NOT NULL
View Code

来看一下最终效果,只有1天的数据,可能看起来不是很直观。

本文的技术点有2个:

  1.利用数字辅助表补全缺失的记录

  2.pivot行转列函数的使用

 

  相关解决方案