Question

我有以下三个表：

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，该输出显示按型号划分的销售总数和按邮政编码排列的销售总数，并按最高值排序。输出应该类似于：

vehicleModel    Sales_By_Model    customerZip   Sales_By_ZIP
        S-10                12          18956              3
   Silverado                10          22789              2
       F-150                 9          12345              2

我已经尝试了以下代码，但我不认为采用这种公牛是正确的方法，因为它没有将vehicleModel与Sales_By_Model的总计数结合起来：

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;

这是我从上面的查询得到的输出：

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

虽然查询确实提取了必要的数据并加入了表，但它并没有组合vehicleModel来增加SALES_BY_MODEL的计数。 customerZIP也是如此，尽管它似乎至少正确地计算了zip。

我的问题是：有没有办法查询这些表格的结果，结合了vehicleModel和customerZip的计数？如果是这样，我该怎么做？我知道通过显式调用JOIN，我上面使用的查询不是完全正确的SQL语法。我的目标是在ONE SELECT Query中完成此操作。

提前感谢您的指导/帮助！

编辑：以下几行样本输入数据：

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我未在原始帖子中列出的其他属性的输入（我在其中显示了表格创建），因为这些属性中的大多数与此处的目标不相关。

Answer 1

实现此目的的一种示例方式表明，您实际上正在运行两个独立的查询，并且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，阻止代码重用，并且通常会违反我能想到的所有相关软件工程原理。

所以，这里有一个非常糟糕的事情......

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)

Oracle SQL连接三个表并按最高值排序

1 个答案: