Question

我有一个查询，可以提取每个日期的订单数量。

SELECT 
    name, CONVERT(VARCHAR(10), order_date, 120) AS order_date, 
    COUNT(1) AS orders    
FROM 
    orders AS od
WHERE 
    id = 73
GROUP BY 
    CONVERT(VARCHAR(10), order_date, 120), name 
ORDER BY 
    order_date, name

以下是我运行查询时得到的结果：

    name    order_date  orders
    --------------------------
    20pmam  2016-07-27  39
    20pmam  2016-07-28  30
    20pmam  2016-07-29  32
    20pmam  2016-07-31  468
    20pmam  2016-08-02  75
    20pmam  2016-07-05  30

我需要我的结果是这样的，新的列日

name    order_date  orders  day 
-------------------------------
20pmam  2016-07-27  39      1
20pmam  2016-07-28  30      2 // days between 2016-07-27 to 2016-07-28
20pmam  2016-07-29  32      3 // days between 2016-07-27 to 2016-07-29
20pmam  2016-07-31  468     5 // days between 2016-07-27 to 2016-07-31
20pmam  2016-08-02  75      7 // days between 2016-07-27 to 2016-08-02
20pmam  2016-08-05  30      10 // days between 2016-07-27 to 2016-08-05

第一个/最小order_date应该被视为第1天（在上面的结果中2016-07-27是第1天）并且应该根据第一个/最小order_date计算其他。

这很容易吗？

我不知道如何获得理想的结果。我很感激任何建议。

Answer 1

您可以执行此操作cross apply以获取每个order_date之前的最短日期，并在datediff中使用它。

SELECT name,CONVERT(VARCHAR(10), order_date, 120) AS order_date, Count(1)  [orders], 
1+coalesce(datediff(day,t.min_date,od.order_date),0) as [Day]
FROM orders AS od
cross apply (select min(od1.order_date) as min_date
             from orders od1 
             where od.id=od1.id and od.name=od1.name and od1.order_date<od.order_date) t
WHERE id = 73
GROUP BY CONVERT(VARCHAR(10), order_date, 120),name,datediff(day,t.min_date,od.order_date)
ORDER BY order_date,name

Answer 2

尝试类似：

SELECT name,
       CONVERT(VARCHAR(10), order_date, 120) AS order_date, 
       Count(1) AS orders,
       DATEDIFF(DAY, first_order_date, order_date) + 1
FROM orders AS od 
JOIN (SELECT min(order_date) AS first_order_date
        FROM orders) as fod ON 1 = 1
WHERE id = 73
GROUP BY CONVERT(VARCHAR(10), order_date, 120),
         name,
         DATEDIFF(DAY, first_order_date, order_date) + 1
ORDER BY order_date, 
         name

希望这能解决您的问题

SQL Server：每日日期与日期列的订单数

2 个答案: