週単位でデータを表示する

Question

私はテーブルに日付の列を持っています。最後の6週間の注文数をWeek1、Week2 ... Week6（これは週番号ではない単純なシリアル番号）として計算する必要があります。ユーザーが2017年12月10日（DD/MM/YYYY）として日付を選択した場合たとえば、それは順序が誰かが私はこのかどうか知らせてくださいでき

として週の日付カウント計算する必要がありますSQLで可能ですか？

Answer 1

これはSQLコードで可能です。ここで

クエリ

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 

Answer 2

は、同じ結果を達成するための別の方法である -

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))