mysql顯示週記錄

Question

嗨我有一個訂單表我想運行一個查詢，將返回在過去4周沒有訂單，如果沒有訂單的任何一週它應該說0順序我正在做的是不工作

ID | order_Date | amount 
1 | 2011-03-01 | 10 
2 | 2011-03-01 | 50 
3 | 2011-02-24 | 60 

select sum(amount) as total from orders group by WEEK(order_date,INTERVAL 3 Week) 

感謝

Answer 1

select monday,monday + interval 6 day as sunday,coalesce(a.total,0) as total from (
select curdate() - 
interval weekday(curdate()) day - interval tmp.digit * 1 week as monday    
from (
select 0 as digit union all 
select 1 union all 
select 2 union all 
select 3 
) as tmp) as t 
left join ( select 
     week(order_date) as nweek, 
     sum(amount) as total 
     from orders 
     where order_date 
     between 
     curdate() - interval weekday(curdate()) day - interval 3 week and 
     curdate() + interval 6 - weekday(curdate()) day 
     group by nweek order by null 
     ) as a 
on week(monday) = a.nweek 
group by a.nweek 
order by monday desc 

結果：

+------------+------------+--------+ 
| monday  | sunday  | total | 
+------------+------------+--------+ 
| 2011-02-28 | 2011-03-06 | 834312 | 
| 2011-02-21 | 2011-02-27 | 818334 | 
| 2011-02-14 | 2011-02-20 | 824032 | 
| 2011-02-07 | 2011-02-13 | 695021 | 
+------------+------------+--------+ 

Answer 2

讓我們看一下你使用的語言：

，將不返回訂單在過去4個星期

在這裏，我們宣佈其下的行應努力總和計數的條件。這些條件在sql中表達爲「where」子句。所以在這種情況下，我們的「where」子句應該比較訂單的日期，並且看到它在一定的時間間隔內。

現在的問題是，如果您的案例中的4周是指「本週和本週之前的三週」，或者「今天和今天之前的27天」。使用「27天」推理是最簡單的，所以我會使用它！因此，我們的標準是訂單日期（在列order_Date中找到）應該在（或大於）之後的（即今天27天前的一天）。這意味着我們需要計算在這一天前27天的一天。我們用SUBDATE和CURDATE函數來做到這一點。這裏只有「where」子句：

WHERE 
    order_Date > SUBDATE(CURDATE(), INTERVAL 27 DAY) 

這樣我們可以找到所有我們想要加在一起的訂單。這裏是列出我們想要的所有訂單的查詢：

SELECT 
    * 
FROM 
    orders 
WHERE 
    order_Date > SUBDATE(CURDATE(), INTERVAL 27 DAY); 

現在我們需要將它們添加到一個結果行中！要做到這一點，我們需要將結果行分組到一個，然後總結所有金額。但「分組依據」僅將所選列分組的行具有相同的值。並且不能保證行中會有這樣的列。所以我們將在查詢的SELECT部分​​創建一個。

SELECT 
    amount, 
    1 AS column_to_group_by 
FROM 
    orders 
WHERE 
    order_Date > SUBDATE(CURDATE(), INTERVAL 27 DAY); 

現在我們可以按列column_to_group_by分組。

SELECT 
    amount, 
    1 AS column_to_group_by 
FROM 
    orders 
WHERE 
    order_Date > SUBDATE(CURDATE(), INTERVAL 27 DAY) 
GROUP BY 
    column_to_group_by; 

結果是現在完全無用的，因爲它的唯一表示該表的第一行的量。但我們現在可以總結所有金額，這樣就可以得到我們想要的答案！

SELECT 
    SUM(amount), 
    1 AS column_to_group_by 
FROM 
    orders 
WHERE 
    order_Date > SUBDATE(CURDATE(), INTERVAL 27 DAY) 
GROUP BY 
    column_to_group_by; 

我希望這個解釋將幫助您採取解決問題的方法應用在這裏:)

Answer 3

select sum(amount) as total from orders 
where 
order_date between date_sub(now(), interval 4 week) and 
date_sub(now(), interval 3 week) 
union 
select sum(amount) as total from orders 
where 
order_date between date_sub(now(), interval 3 week) and 
date_sub(now(), interval 2 week) 
union 
select sum(amount) as total from orders 
where 
order_date between date_sub(now(), interval 2 week) and 
date_sub(now(), interval 1 week) 
union 
select sum(amount) as total from orders 
where 
order_date between date_sub(now(), interval 1 week) and now();