我有以下三张表:
CREATE TABLE CUSTOMERS
(customerID INT PRIMARY KEY,
customerZip VARCHAR(15) NOT NULL);
CREATE TABLE VEHICLES
(vehicleVIN VARCHAR(25) PRIMARY KEY,
vehicleModel VARCHAR(15) NOT NULL);
CREATE TABLE SALES
(saleID INT PRIMARY KEY,
customerID INT,
vehicleVIN VARCHAR(25),
CONSTRAINT SALES_FK1 FOREIGN KEY (customerID) REFERENCES CUSTOMERS(customerID),
CONSTRAINT SALES_FK2 FOREIGN KEY (vehicleVIN) REFERENCES VEHICLES(vehicleVIN));我正在尝试开发一个查询来连接这三个表,并生成一个输出,该输出按模型显示销售总量,并按邮政编码显示销售总量,首先按最高值排序。输出应该类似于以下内容:
vehicleModel Sales_By_Model customerZip Sales_By_ZIP
S-10 12 18956 3
Silverado 10 22789 2
F-150 9 12345 2我尝试了以下代码,但我不认为这是正确的方法,因为它没有将vehicleModel与Sales_By_Model的总数组合在一起:
SELECT DISTINCT v.vehicleModel, COUNT(*) OVER (PARTITION BY s.vehicleVIN) "SALES_BY_MODEL", c.customerZip, COUNT(*) OVER (PARTITION BY c.customerZip )"SALES_BY_ZIP"
FROM SALES s, VEHICLES v, CUSTOMERS c
WHERE s. vehicleVIN = v. vehicleVIN
and c. customerID = s. customerID
ORDER BY 2 DESC , 4 DESC;这是我从上面所做的查询中得到的输出:
VEHICLEMODEL SALES_BY_MODEL CUSTOMERZIP SALES_BY_ZIP
accord 1 89523 6
altima 1 89523 6
escalade 1 89523 6
f-150 1 89523 6
impala 1 89523 6尽管查询确实会提取必要的数据并连接表,但它不会组合vehicleModel(s)来增加SALES_BY_MODEL的计数。同样的情况也发生在customerZIP上,尽管它似乎至少正确地计算了拉链。
我的问题是:是否有一种方法可以将vehicleModel和customerZip的计数结合起来查询这些表?如果是的话,我该怎么做呢?通过显式调用JOIN,我知道上面使用的查询不是完全正确的SQL语法。我的目标是在一个选择查询中完成这个任务。
提前感谢您的指导/协助!
编辑:下面是几行示例输入数据:
INSERT INTO CUSTOMERS
(customerID,customerFirName,customerLasName,customerMiName,customerStreet,customerState,customerCity,customerZip)
VALUES
(1,'Steven','Christman','J','11111 Address Way','Maryland','Hollywood','20636');
INSERT INTO CUSTOMERS
(customerID,customerFirName,customerLasName,customerMiName,customerStreet,customerState,customerCity,customerZip)
VALUES
(2,'Bob','Seagram','A','22222 Seagram Lane','Texas','Houston','77001');
INSERT INTO CUSTOMERS
(customerID,customerFirName,customerLasName,customerMiName,customerStreet,customerState,customerCity,customerZip)
VALUES
(3,'Sally','Anderson','P','33333 Pheonix Drive','Arizona','Pheonix','85001');
INSERT INTO VEHICLES
(vehicleVIN,vehicleType,vehicleMake,vehicleModel,vehicleWhereFrom,vehicleWholesaleCost,vehicleTradeID)
VALUES
('147258HHE91K3RT','compact','chevrolet','spark','Maryland',20583.00,NULL);
INSERT INTO VEHICLES
(vehicleVIN,vehicleType,vehicleMake,vehicleModel,vehicleWhereFrom,vehicleWholesaleCost,vehicleTradeID)
VALUES
('789456ERT0923RFB6','Midsize','ford','Taurus','washington, d.c.',25897.22,1);
INSERT INTO VEHICLES
(vehicleVIN,vehicleType,vehicleMake,vehicleModel,vehicleWhereFrom,vehicleWholesaleCost,vehicleTradeID)
VALUES
('1234567890QWERTYUIOP','fullsize','Lincoln','towncar','Virginia',44222.10,NULL);
INSERT INTO SALES
(saleID,grossSalePrice,vehicleStatus,saleDate,saleMileage,customerID,salespersonID,vehicleVIN)
VALUES
(1,25987.28,'sold',date '2012-10-15',10,1,1,'147258HHE91K3RT');
INSERT INTO SALES
(saleID,grossSalePrice,vehicleStatus,saleDate,saleMileage,customerID,salespersonID,vehicleVIN)
VALUES
(2,29999.99,'sold',date '2012-10-17',50087,2,2,'789456ERT0923RFB6');
INSERT INTO SALES
(saleID,grossSalePrice,vehicleStatus,saleDate,saleMileage,customerID,salespersonID,vehicleVIN)
VALUES
(3,47490.88,'sold',date '2012-11-05',30,3,3,'1234567890QWERTYUIOP');正如您可能注意到的,上面的输入数据反映了一些其他属性的输入,而我在最初的文章中没有列出这些属性(我在其中显示了表创建),因为这些属性中的大多数与这里的目标无关。
发布于 2018-01-30 14:55:25
实现这一点的一个示例方法表明,您实际上正在运行两个独立的查询,只是假装它们是相关的。
这样做会增加执行成本,降低测试和/或维护的能力,防止代码重用,并且通常违背我能想到的每一个相关的软件工程原则。
所以,这是如何做一件非常糟糕的事情..。
WITH
ranked_model_sales AS
(
SELECT
vehicles.vehicleModel,
COUNT(*) AS total_model_sales,
ROW_NUMBER() OVER (ORDER BY COUNT(*) DESC) AS rank_id
FROM
sales
INNER JOIN
vehicles
ON sales.vehicleVIN = vehicles.vehicleVIN
GROUP BY
vehicles.vehicleModel
),
ranked_zip_sales AS
(
SELECT
customers.customerZip,
COUNT(*) AS total_zip_sales,
ROW_NUMBER() OVER (ORDER BY COUNT(*) DESC) AS rank_id
FROM
sales
INNER JOIN
customers
ON sales.customerID= customers.customerID
GROUP BY
customers.customerZip
)
SELECT
m.vehicleModel,
m.total_model_sales,
z.customerZip,
z.total_zip_sales
FROM
ranked_model_sales m
FULL OUTER JOIN
ranked_zip_sales z
ON m.rank_id = z.rank_id
ORDER BY
COALESCE(m.rank_id, z.rank_id)https://stackoverflow.com/questions/48523731
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