Question

我有一个简单的表格，其中包含某个组的日期和总销售额：

date       | totalsales
=======================
2014-05-01 | 3000
2014-05-02 | 3100
2014-05-03 | 3500
2014-05-04 | 3650

我喜欢计算一些东西：

每日销售
平均销售额
增长％

结果应该看起来像（手动计算所以可能错误:)）

date       | sales  | average | growth
=======================================
2014-05-01 |   0    |  0      |   0
2014-05-02 | 100    | 50      | 100
2014-05-03 | 400    | 166.66  | 400
2014-05-04 | 150    | 162.5   |  37.5

这是否可以在sql语句中使用，还是应该用PHP或其他服务器软件计算？

Answer 1

假设每个日期都有自己唯一的行，您可以通过加入原来的表来完成：

SELECT t1.Date, CASE WHEN t2.Date IS NULL THEN 0 ELSE (t1.totalsales - t2.totalsales)
END AS sales
FROM table t1
LEFT JOIN table t2 ON t2.Date = DATE_ADD(t1.Date, INTERVAL -1 DAY)
ORDER BY 1

这至少会给你第一列，你应该能够从那里找出其余的数学。在此处使用左连接和CASE语句非常重要，否则您将无法获得表格中的最低日期（第一行）

如果每个日期不获得自己的唯一行，则此方法仍然可行，您只需要在日期列上使用GROUP BY和SUM在子查询中创建数据集。

Answer 2

以下是每行没有子选择的完整查询:(感谢 @nmarsh 撰写最难的部分）

请参阅SQL小提琴： http://sqlfiddle.com/#!2/be4654/34/0

SELECT 
  t1.Date,
  CASE
    WHEN t2.date IS NULL THEN 0 ELSE (t1.totalSales - t2.totalSales)
    END AS sales,
  CASE 
    WHEN t2.date IS NULL THEN 0 / (@curRow := @curRow + 1) ELSE ((@curSum := @curSum + (t1.totalSales - t2.totalSales)) / (@curRow := @curRow + 1))
    END AS average,
  CASE
    WHEN t3.date IS NULL AND t2.date IS NULL THEN 0
    WHEN t3.date IS NULL THEN (t1.totalSales - t2.totalSales)
    WHEN t2.date IS NULL THEN 0 ELSE ((t1.totalSales - t2.totalSales) * 100) / (t2.totalSales - t3.totalSales)
    END AS growth
FROM test t1
LEFT JOIN test t2 ON t2.date = DATE_ADD(t1.Date, INTERVAL -1 DAY)
LEFT JOIN test t3 ON t3.date = DATE_ADD(t2.Date, INTERVAL -1 DAY)
JOIN (SELECT @curRow := 0) r
JOIN (SELECT @curSum := 0) ct
ORDER BY 1;

原始表：

date       | totalsales
=======================
2014-05-01 |   3000
2014-05-02 |   3100
2014-05-03 |   3500
2014-05-04 |   3650

<强>输出

date       | sales  | average | growth
=======================================
2014-05-01 |   0    |  0      | 0
2014-05-02 | 100    | 50      | 100
2014-05-03 | 400    | 166.66  | 400
2014-05-04 | 150    | 162.5   | 37.5

Answer 3

您可以使用recursive statement。在每次迭代中，计算一天所请求的数据，并删除第一天（最早）。

你也可以使用PHP，这看起来更好，因为你不想在MySQL表上加载太多负载以防万一它不能节省你的时间/计算。

Answer 4

对数字不太确定，但如果您对结果更具体，我可以仔细检查。

您可以使用ROW_NUMBER（）创建2个数据集并将它们连接到ROW_NUMBER（）和ROW_NUMBER（） - 1以获取当前值和先前值的偏移量以计算增长。样本：

DECLARE @Data TABLE (SalesDate DATETIME, totalSales INT)

INSERT INTO @Data (SalesDate , totalSales) VALUES ('2014-05-01' , 3000)
INSERT INTO @Data (SalesDate , totalSales) VALUES ('2014-05-02' , 3100)
INSERT INTO @Data (SalesDate , totalSales) VALUES ('2014-05-03' , 3500)
INSERT INTO @Data (SalesDate , totalSales) VALUES ('2014-05-04' , 3650)


SELECT  
      CurrentDt.SalesDate 
     ,ISNULL(CurrentDt.totalSales - PreviousDt.totalSales ,0)                       AS  Sales
     ,FirstDate.FirstDate
     , NULLIF(CAST((CurrentDt.SalesDate - FirstDate.FirstDate) AS INT)+1,0)         AS SellingDays
     ,(ISNULL(CurrentDt.totalSales - PreviousDt.totalSales ,0)) 
     / NULLIF(CAST((CurrentDt.SalesDate - FirstDate.FirstDate) AS INT)+1,0)         AS  AverageSales
FROM

    (SELECT Min(SalesDate) AS FirstDate FROM @Data) AS FirstDate,
    /*Base Sales Data*/
    (
    SELECT
         ROW_NUMBER() OVER(ORDER BY SalesDate) AS RowNum
        ,SalesDate 
        ,totalSales
    FROM
        @Data 
    ) AS CurrentDt

    /*Previous Value for Growth*/
    LEFT JOIN 
    (
    SELECT 
         ROW_NUMBER() OVER(ORDER BY SalesDate) AS RowNum
        ,SalesDate 
        ,totalSales  
    FROM
        @Data 
    ) AS PreviousDt
        ON CurrentDt.RowNum  -1 = PreviousDt.RowNum

我使用过MSSQL，但MySQL支持ROW_NUMBER OVER。

Answer 5

我希望此查询可以帮助您

SELECT 
    sample.id,
    sample.date AS oggi,
    sample.value AS sales,
    ((SELECT SUM(sample.value) FROM sample WHERE  sample.date <= oggi    ) / (SELECT COUNT(sample.value) FROM sample WHERE  sample.date <= oggi    ) ) AS avarege,  
    sample.value / IF((SELECT sample.value FROM sample WHERE  sample.date =  (oggi - INTERVAL 1 DAY )) = 0,sample.value,(SELECT sample.value FROM sample WHERE  sample.date =  (oggi - INTERVAL 1 DAY ))) *100 AS 'growt-percent'
    -- (SELECT SUM(sample.value) FROM sample WHERE  sample.date <= oggi    ) AS somma,
    -- (SELECT count(sample.value) FROM sample WHERE  sample.date <= oggi    ) AS conta,
    -- (SELECT sample.value FROM sample WHERE  sample.date =  (oggi - INTERVAL 1 DAY )) as valoreieri,
FROM sample 
WHERE sample.date BETWEEN '2014-05-01 00:00:00' AND '2014-05-31 00:00:00'

表数据是

id日期值

1  2014-05-01 00:00:00         0
2  2014-05-02 00:00:00       100
3  2014-05-03 00:00:00       400
4  2014-05-04 00:00:00       150
5  2014-05-05 00:00:00       200

结果是

id oggi sales avarege growt-percent

1  2014-05-01 00:00:00       0  0.0000    (NULL)         
2  2014-05-02 00:00:00     100  50.0000   100.0000       
3  2014-05-03 00:00:00     400  166.6667  400.0000       
4  2014-05-04 00:00:00     150  162.5000  37.5000        
5  2014-05-05 00:00:00     200  170.0000  133.3333

请注意我使用datetime字段不仅仅是日期如果您对查询有疑问，请

抱歉我的英文不好

编辑最后3行被评论，因为我只将它用于测试

计算MySQL中的每日销售额

5 个答案: