以每周为单位显示数据

我在表中有一个日期列。根据选择的日期，它应该计算过去6周的订单数量为Week1，Week2 ... Week6（这不是其中的简单序列号）。例如，如果用户选择了日期2017年12月10日（日/月/年），那么它应该计算订单数为周日期为以每周为单位显示数据

below image:

有人可以请让我知道这是否在SQL中可能吗？

来源

2017-10-12 Aasish Kumar

此数据不neccessarily属于到数据库。只是在飞行计算，而不是...顺便说一句，摆脱存储' - '分离的数据分贝... –

我该如何计算。我试着用下面的代码'Select SUM（当CMTOrder.OrderPickUpTimestamp CMTOrder.CreateTimestamp和Dateadd（dd，-7，CMTOrder.CreateTimestamp）then 1 else 0 end）没有为我工作。上面的截图我只是举了一个例子。我的数据库没有保存像tat这样的数据。 –

这是可能的SQL代码。

查询

SELECT 
    CONCAT(week1.first_day, '-', week1.second_day) AS Week1 
, CONCAT(week2.first_day, '-', week2.second_day) AS Week2 
, CONCAT(week3.first_day, '-', week3.second_day) AS Week3 
, CONCAT(week4.first_day, '-', week4.second_day) AS Week4 
, CONCAT(week5.first_day, '-', week5.second_day) AS Week5 
, CONCAT(week6.first_day, '-', week6.second_day) AS Week6 
FROM (
SELECT 
    DATE_FORMAT(@first_date_of_week, '%d/%m/%Y') AS first_day 
    , DATE_FORMAT(@date, '%d/%m/%Y') AS second_day 
) 
week1 
CROSS JOIN (
    SELECT 
     DATE_FORMAT(@first_date_of_week - INTERVAL 1 WEEK, '%d/%m/%Y') AS first_day 
    , DATE_FORMAT(@first_date_of_week - INTERVAL 1 WEEK + INTERVAL 1 WEEK - INTERVAL 1 DAY, '%d/%m/%Y') AS second_day 
) 
week2 
CROSS JOIN (
    SELECT 
     DATE_FORMAT(@first_date_of_week - INTERVAL 2 WEEK, '%d/%m/%Y') AS first_day 
    , DATE_FORMAT(@first_date_of_week - INTERVAL 2 WEEK + INTERVAL 1 WEEK - INTERVAL 1 DAY, '%d/%m/%Y') AS second_day 
) 
week3 
CROSS JOIN (
    SELECT 
     DATE_FORMAT(@first_date_of_week - INTERVAL 3 WEEK, '%d/%m/%Y') AS first_day 
    , DATE_FORMAT(@first_date_of_week - INTERVAL 3 WEEK + INTERVAL 1 WEEK - INTERVAL 1 DAY, '%d/%m/%Y') AS second_day 
) 
week4 
CROSS JOIN (
    SELECT 
     DATE_FORMAT(@first_date_of_week - INTERVAL 4 WEEK, '%d/%m/%Y') AS first_day 
    , DATE_FORMAT(@first_date_of_week - INTERVAL 4 WEEK + INTERVAL 1 WEEK - INTERVAL 1 DAY, '%d/%m/%Y') AS second_day 
) 
week5 
CROSS JOIN (
    SELECT 
     DATE_FORMAT(@first_date_of_week - INTERVAL 5 WEEK, '%d/%m/%Y') AS first_day 
    , DATE_FORMAT(@first_date_of_week - INTERVAL 5 WEEK + INTERVAL 1 WEEK - INTERVAL 1 DAY, '%d/%m/%Y') AS second_day 
) 
week6 

CROSS JOIN (
    SELECT 
     @date := STR_TO_DATE('12/10/2017', '%d/%m/%Y') AS DATE 
    , @first_date_of_week := @date - INTERVAL (DAYOFWEEK(@date) - 1) DAY AS first_date_of_the_week 
) init_user_params

结果

Week1     Week2     Week3     Week4     Week5     Week6     
--------------------- --------------------- --------------------- --------------------- --------------------- ----------------------- 
08/10/2017-12/10/2017 01/10/2017-07/10/2017 24/09/2017-30/09/2017 17/09/2017-23/09/2017 10/09/2017-16/09/2017 03/09/2017-09/09/2017

来源

2017-10-12 12:56:00

这里是另一种方式来达到同样的效果 -

select CONCAT(subdate(CURDATE(), INTERVAL (weekday(CURDATE())+1) DAY), ' - ', CURDATE()) 
     ,CONCAT(subdate(date_sub(CURDATE(), interval 7 day), INTERVAL (weekday(CURDATE())+1) DAY), ' - ', subdate(CURDATE(), INTERVAL (weekday(CURDATE())+2) DAY)) 
     ,CONCAT(subdate(date_sub(CURDATE(), interval 14 day), INTERVAL (weekday(CURDATE())+1) DAY), ' - ', subdate(date_sub(CURDATE(), interval 7 day), INTERVAL (weekday(CURDATE())+2) DAY)) 
     ,CONCAT(subdate(date_sub(CURDATE(), interval 21 day), INTERVAL (weekday(CURDATE())+1) DAY), ' - ', subdate(date_sub(CURDATE(), interval 14 day), INTERVAL (weekday(CURDATE())+2) DAY)) 
     ,CONCAT(subdate(date_sub(CURDATE(), interval 28 day), INTERVAL (weekday(CURDATE())+1) DAY), ' - ', subdate(date_sub(CURDATE(), interval 21 day), INTERVAL (weekday(CURDATE())+2) DAY)) 
     ,CONCAT(subdate(date_sub(CURDATE(), interval 35 day), INTERVAL (weekday(CURDATE())+1) DAY), ' - ', subdate(date_sub(CURDATE(), interval 28 day), INTERVAL (weekday(CURDATE())+2) DAY))

来源

2017-10-12 13:10:58

以每周为单位显示数据

回答

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