Question

我需要列出每月显示利润的所有城区。

我不知道如何将产品的计数值与子查询中的月份联系起来。

有人可以帮我一把吗？

SELECT 
       districts.id,
       districts.name,
       mmonth.m10
FROM districts
INNER JOIN
  (SELECT 
       SUM(product_order.price_of_day * product_order.amount) AS m10
   FROM product_order
   INNER JOIN orders ON product_order.order_id = orders.id
   WHERE MONTH(orders.date_time) = 10 ) mmonth
INNER JOIN cities ON cities.id = districts.city_id
WHERE cities.id = 2;

http://sqlfiddle.com/#!2/8fb34/4

看看我需要什么。

| id |     name   | m10 | m11 | m12 |
------------------------------------
|  3 | district 3 | 730 | 360 | 453 |
------------------------------------
|  4 | district 4 | 730 | 250 | 856 |
-------------------------------------

我如何总结这样的月份？

| id |     name   | m10 | m11 | m12 | total |
---------------------------------------------
|  3 | district 3 | 730 | 360 | 453 | 1543  |
---------------------------------------------
|  4 | district 4 | 730 | 250 | 856 | 1836  |
---------------------------------------------

Answer 1

修改

这应该这样做：

SELECT districts.id, districts.name, SUM(IF(MONTH(orders.date_time) = 10, product_order.price_of_day * product_order.amount, 0)) AS m10, SUM(IF(MONTH(orders.date_time) = 11, product_order.price_of_day * product_order.amount, 0)) AS m11, SUM(IF(MONTH(orders.date_time) = 12, product_order.price_of_day * product_order.amount, 0)) AS m12, SUM(product_order.price_of_day * product_order.amount) AS total FROM cities LEFT JOIN districts ON cities.id = districts.city_id LEFT JOIN orders ON orders.district_id = districts.id LEFT JOIN product_order ON product_order.order_id = orders.id WHERE cities.id = 2 GROUP BY districts.id;

Answer 2

您应该使用GROUP BY（请参阅：http://technet.microsoft.com/en-us/library/ms177673.aspx）

SELECT
    [IdentifyingDataColumnName]
    ...
    ,[IdentifyingDataColumnName1]
    ,[CalculatedField] =  SUM(<column name(/s) / Calculation to sum>)
FROM
    Table 1
    LEFT JOIN 2
GROUP BY
    [IdentifyingDataColumnName]
    ...
    ,[IdentifyingDataColumnName1]
HAVING  
    SUM(<column name(/s) / Calculation to sum>) > 0

重新阅读你的问题（并意识到我完全误读了它！）你可以做到以下几点：

SELECT 
    districts.id
    ,districts.name
    ,SUM(CASE WHEN MONTH(orders.date_time) = 10 THEN (product_order.price_of_day * product_order.amount) ELSE 0 END) AS m10
    ,SUM(CASE WHEN MONTH(orders.date_time) = 11 THEN (product_order.price_of_day * product_order.amount) ELSE 0 END) AS m11
    ,SUM(CASE WHEN MONTH(orders.date_time) = 12 THEN (product_order.price_of_day * product_order.amount) ELSE 0 END) AS m12
FROM 
    districts 
    INNER JOIN cities ON cities.id = districts.city_id
    INNER JOIN orders ON districts.id = orders.district_id
    INNER JOIN product_order ON product_order.order_id = orders.id
WHERE 
    cities.id = 2
GROUP BY 
    districts.id
    ,districts.name

添加总列数并不容易，只需删除CASE... WHEN... THEN... ELSE... END即可将它们全部加起来;下面的代码也被修改为使用LEFT JOIN，以确保如果没有订单的城市存在区域，它仍然会列出月值为0（有关详细信息，请参阅：http://technet.microsoft.com/en-us/library/aa178429(v=SQL.80).aspx ）

SELECT 
    districts.id
    ,districts.name
    ,SUM(CASE WHEN MONTH(orders.date_time) = 10 THEN (product_order.price_of_day * product_order.amount) ELSE 0 END) AS m10
    ,SUM(CASE WHEN MONTH(orders.date_time) = 11 THEN (product_order.price_of_day * product_order.amount) ELSE 0 END) AS m11
    ,SUM(CASE WHEN MONTH(orders.date_time) = 12 THEN (product_order.price_of_day * product_order.amount) ELSE 0 END) AS m12
    ,SUM(product_order.price_of_day * product_order.amount) AS Total
FROM 
    districts 
    INNER JOIN cities ON cities.id = districts.city_id
    LEFT JOIN orders ON districts.id = orders.district_id
    LEFT JOIN product_order ON product_order.order_id = orders.id
WHERE 
    cities.id = 2
GROUP BY 
    districts.id
    ,districts.name

如何列出每月显示利润的城市的所有地区

2 个答案: