城市客户总销售额

Question

我有以下 SQL 语句，显示至少完成 2 个订单的城市客户的总销售额。但是假设我只想显示某人至少完成了2个订单的城市，并且有两个客户住在这个城市/城市，所以我想要做的是挑选城市/城市一个客户至少下了2个订单，然后显示两个居住在这个城市的客户的总销售额，即使另一个客户只下了一个订单，那是怎么做的，是否有COUNT的比较语句- 操作能够显示城市中所有客户的销售额，如果是的话 - 它是如何表示的？

SELECT c.CityName, SUM(p.Price * o2.Orderquantity) AS 'TotalSalesAmount'
FROM Customers c, Order1 o1, Orderrader o2, Products p,
(SELECT o1.CustomerNr 
FROM Order1 o1
GROUP BY o1.CustomerNr 
HAVING COUNT(o1.CustomerNr) >= 2)
AS a WHERE c.CustomerNr = a.CustomerNr AND c.CustomerNr = o1.CustomerNr
AND o1.Ordernr = o2.Ordernr AND o2.Productnr = p.Productnr
GROUP BY c.CityName

我使用的表结构如下：

'Customers' has the columns: CustomerNr, City Name
'Order1' has columns: Ordernr, CustomerNr
'Order2' has columns: Ordernr, Productnr, Order quantity
'Products' has columns: ProductNr, Price

示例数据：“客户”表的示例数据：

- CustomerNr  CityName:
- 01109       New York
- 01999       Los Angeles
- 20090       New York

“订单 1”的示例数据：

- Ordernr  CustomerNr
- 1306     01109
- 1307     01109
- 1308     20090

“订单 2”的示例数据：

- OrderNr   ProductNr  Order quantity:
- 1306      15-116     3
- 1306      46-701     2
- 1307      15-96      1
- 1308      17-91      1

（等等...）

“产品”的示例数据：

- ProductNr  Price:
- 15-116     44.00
- 15-96      28.50
- 46-701     3000.00
- 17-91      200.00

等等……

根据上面的SQL语句和示例数据，我想要的结果是：

- CityName  TotalSalesAmount:
- New York   6360.50

Answer 1

尝试使用子查询。我知道这看起来有点讨厌，但肯定是有效的。

事实证明，在您的 Order1 示例中，相同的 CustomerNr 有一个重复的 Ordernr (1306)。我假设您的真实数据可能并非如此。

SELECT c.CityName, 
(Select SUM(order2.quantity * products.Price) from order1 
INNER JOIN Customers On Customers.CustomerNr=order1.CustomerNr 
INNER JOIN Order2 ON Order2.Ordernr=Order1.Ordernr 
INNER JOIN Products ON Products.ProductNr=Order2.ProductNr 
WHERE Customers.CityName=c.CityName) AS 'TotalSalesAmount'
FROM Order1 o1
INNER JOIN (SELECT o1.CustomerNr 
        FROM Order1 o1 
        GROUP BY o1.CustomerNr 
        HAVING COUNT(o1.CustomerNr) >= 2
       ) AS a ON o1.CustomerNr = a.CustomerNr 
INNER JOIN Order2 o2 ON o1.Ordernr = o2.Ordernr 
INNER JOIN Customers c ON o1.CustomerNr = c.CustomerNr
INNER JOIN Products P ON o2.ProductNr = P.ProductNr
GROUP BY c.CityName;

Answer 2

试一试。

SELECT CityName, SUM(quantity * Price) AS TotalSalesAmount
FROM (
    SELECT c.CityName, p.ProductNr, b.quantity, p.Price, (b.quantity * p.Price) as total --, SUM(b.quantity * p.Price) AS TotalSalesAmount
    FROM customers c
        INNER JOIN order1 a ON c.CustomerNr = a.CustomerNr
        INNER JOIN order2 b ON a.Ordernr = b.Ordernr
        INNER JOIN products p ON b.Productnr = p.ProductNr
    GROUP BY c.CityName, p.ProductNr, b.quantity, p.Price
    ) AS cust
GROUP BY CityName
HAVING COUNT(cityname) >= 2 

Answer 3

试试这个。

SELECT c.CityName,
       Sum(o2.Orderquantity * p.Price) Total_sale
FROM   Customers C
       JOIN (SELECT o.CustomerNr
             FROM   Order1 o
             GROUP  BY o.CustomerNr
             HAVING Count(o.CustomerNr) >= 2) su
         ON c.CustomerNr = su.CustomerNr
       JOIN Customers c1
         ON c1.CityName = c.CityName
       JOIN Order1 o1
         ON o1.CustomerNr = c1.CustomerNr
       JOIN Order2 o2
         ON o2.Ordernr = o1.Ordernr
       JOIN Products P
         ON o2.ProductNr = P.ProductNr
GROUP  BY c.CityName   

Answer 4

我已将您的问题解释为“我需要按城市列出的所有销售额的总和，对于单个客户下过多个订单的任何城市。”

select 
    c.City,
    SUM(p.Price * o2.Orderquantity) AS 'TotalSalesAmount' 
from
    (select c.City
    from @t_Customers c
    inner join @t_Order1 o1
        on o1.CustomerNr = c.CustomerNr
    group by c.City
    having count(c.City) > 1) as ct
    inner join @t_Customers c
        on c.City = ct.City
    inner join @t_Order1 o1
        on o1.CustomerNr = c.CustomerNr
    inner join @t_Order2 o2
        on o2.OrderNr = o1.Ordernr
    inner join @t_Products p
        on p.ProductNr = o2.ProductNr
group by c.City

Answer 5

select ci.city_name,pr.product_name,round(sum(ii.line_total_price),2) as tot 
from city ci ,customer cu,invoice i,invoice_item ii,product pr where ci.id=cu.city_id and cu.id=i.customer_id and i.id=ii.invoice_id and ii.product_id=pr.id
group by ci.city_name,pr.product_name
order by tot desc,ci.city_name, pr.product_name