以每週爲單位顯示數據

我在表中有一個日期列。根據選擇的日期，它應該計算過去6周的訂單數量爲Week1，Week2 ... Week6（這不是其中的簡單序列號）。例如，如果用戶選擇了日期2017年12月10日（日/月/年），那麼它應該計算訂單數爲週日期爲以每週爲單位顯示數據

below image:

有人可以請讓我知道這是否在SQL中可能嗎？

來源

2017-10-12 Aasish Kumar

此數據不neccessarily屬於到數據庫。只是在飛行計算，而不是...順便說一句，擺脫存儲' - '分離的數據分貝... –

我該如何計算。我試着用下面的代碼'Select SUM（當CMTOrder.OrderPickUpTimestamp CMTOrder.CreateTimestamp和Dateadd（dd，-7，CMTOrder.CreateTimestamp）then 1 else 0 end）沒有爲我工作。上面的截圖我只是舉了一個例子。我的數據庫沒有保存像tat這樣的數據。 –

這是可能的SQL代碼。

查詢

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

結果

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

來源

2017-10-12 12:56:00

這裏是另一種方式來達到同樣的效果 -

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))

來源

2017-10-12 13:10:58

以每週爲單位顯示數據

回答

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