SQL 查询基础(2)-实例
时间:2014-07-22 23:17:15
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上次说到SQL查询语句的逻辑执行过程,现在来用一个实例说明一下逻辑执行的过程。
前提:我们有三个表,分别记住客户信息、订单信息和产品信息
客户信息表:Customer
ID,Name,Adress,PhoneNumber
ID | Name | Adress | PhoneNumber |
1 | CompanyA | No.1 Street | 123456 |
2 | CompanyB | No.2 Street | 23453 |
3 | CompanyC | No.3 Street | 45321 |
4 | CompanyD | No.4 Street | 4567890 |
5 | CompanyE | No.5 Street | 123890 |
订单信息表:OrderInfo
ID,Cus_ID,Product_ID,OrderDate,Qua
ID | Cus_ID | Product_ID | OrderDate | Qua |
1 | 2 | 2 | 1/1/2012 | 1000 |
2 | 4 | 1 | 5/2/2012 | 500 |
3 | 1 | 5 | 9/10/2013 | 3069 |
产品信息表:Product
ID,Name,Price
ID | Name | Price |
1 | ABC | 10 |
2 | DEF | 16 |
3 | GHI | 18 |
4 | JKL | 9 |
5 | MNO | 100 |
1 SELECT 2 C.Name AS CustomerName, 3 O.OrderDate AS OrderDate, 4 O.Qua, 5 P.Name AS ProductName, 6 P.Price AS Price 7 FROM Customer C 8 INNER JOIN OrderInfo O 9 ON C.ID = O.Cus_ID 10 INNER JOIN Product P 11 ON O.Product_ID = P.ID
现在我们根据上一篇文章中的逻辑执行顺序分析一下查询的结果。
1) FROM Customer 表作为主表 T1
2) JOIN OrderInfo 表作为表 T2,这时将 T1和T2做笛卡尔乘积。得到的虚拟表如下:
ID | Name | Adress | PhoneNumber | ID | Cus_ID | Product_ID | OrderDate | Qua |
1 | CompanyA | No.1 Street | 123456 | 1 | 2 | 2 | 1/1/2012 | 1000 |
2 | CompanyB | No.2 Street | 23453 | 1 | 2 | 2 | 1/1/2012 | 1000 |
3 | CompanyC | No.3 Street | 45321 | 1 | 2 | 2 | 1/1/2012 | 1000 |
4 | CompanyD | No.4 Street | 4567890 | 1 | 2 | 2 | 1/1/2012 | 1000 |
5 | CompanyE | No.5 Street | 123890 | 1 | 2 | 2 | 1/1/2012 | 1000 |
1 | CompanyA | No.1 Street | 123456 | 2 | 4 | 1 | 5/2/2012 | 500 |
2 | CompanyB | No.2 Street | 23453 | 2 | 4 | 1 | 5/2/2012 | 500 |
3 | CompanyC | No.3 Street | 45321 | 2 | 4 | 1 | 5/2/2012 | 500 |
4 | CompanyD | No.4 Street | 4567890 | 2 | 4 | 1 | 5/2/2012 | 500 |
5 | CompanyE | No.5 Street | 123890 | 2 | 4 | 1 | 5/2/2012 | 500 |
1 | CompanyA | No.1 Street | 123456 | 3 | 1 | 5 | 9/10/2013 | 3069 |
2 | CompanyB | No.2 Street | 23453 | 3 | 1 | 5 | 9/10/2013 | 3069 |
3 | CompanyC | No.3 Street | 45321 | 3 | 1 | 5 | 9/10/2013 | 3069 |
4 | CompanyD | No.4 Street | 4567890 | 3 | 1 | 5 | 9/10/2013 | 3069 |
5 | CompanyE | No.5 Street | 123890 | 3 | 1 | 5 | 9/10/2013 | 3069 |
执行 ON 后边的条件:C.ID = O.Cus_ID,保留结果为 TRUE 的记录,结果如下:
ID | Name | Adress | PhoneNumber | ID | Cus_ID | Product_ID | OrderDate | Qua | Result |
1 | CompanyA | No.1 Street | 123456 | 1 | 2 | 2 | 1/1/2012 | 1000 | FALSE |
2 | CompanyB | No.2 Street | 23453 | 1 | 2 | 2 | 1/1/2012 | 1000 | TRUE |
3 | CompanyC | No.3 Street | 45321 | 1 | 2 | 2 | 1/1/2012 | 1000 | FALSE |
4 | CompanyD | No.4 Street | 4567890 | 1 | 2 | 2 | 1/1/2012 | 1000 | FALSE |
5 | CompanyE | No.5 Street | 123890 | 1 | 2 | 2 | 1/1/2012 | 1000 | FALSE |
1 | CompanyA | No.1 Street | 123456 | 2 | 4 | 1 | 5/2/2012 | 500 | FALSE |
2 | CompanyB | No.2 Street | 23453 | 2 | 4 | 1 | 5/2/2012 | 500 | FALSE |
3 | CompanyC | No.3 Street | 45321 | 2 | 4 | 1 | 5/2/2012 | 500 | FALSE |
4 | CompanyD | No.4 Street | 4567890 | 2 | 4 | 1 | 5/2/2012 | 500 | TRUE |
5 | CompanyE | No.5 Street | 123890 | 2 | 4 | 1 | 5/2/2012 | 500 | FALSE |
1 | CompanyA | No.1 Street | 123456 | 3 | 1 | 5 | 9/10/2013 | 3069 | TRUE |
2 | CompanyB | No.2 Street | 23453 | 3 | 1 | 5 | 9/10/2013 | 3069 | FALSE |
3 | CompanyC | No.3 Street | 45321 | 3 | 1 | 5 | 9/10/2013 | 3069 | FALSE |
4 | CompanyD | No.4 Street | 4567890 | 3 | 1 | 5 | 9/10/2013 | 3069 | FALSE |
5 | CompanyE | No.5 Street | 123890 | 3 | 1 | 5 | 9/10/2013 | 3069 | FALSE |
由于用的是 INNER JOIN,所以不考虑 T1(T2)中存在而T2(T1)中不存在的情况,继续执行笛卡尔乘积和 ON 条件筛选,得到结果如下表:
ID | Name | Adress | PhoneNumber | ID | Cus_ID | Product_ID | OrderDate | Qua | ID | Name | Price |
4 | CompanyD | No.4 Street | 4567890 | 2 | 4 | 1 | 5/2/2012 | 500 | 1 | ABC | 10 |
2 | CompanyB | No.2 Street | 23453 | 1 | 2 | 2 | 1/1/2012 | 1000 | 2 | DEF | 16 |
1 | CompanyA | No.1 Street | 123456 | 3 | 1 | 5 | 9/10/2013 | 3069 | 5 | MNO | 100 |
由于没有用到 WHERE 条件,没有 GROUP BY,没有 HAVING 之类的操作。直接执行 SELECT 中需要选中的列,返回数据查询结果。
CustomerName | OrderDate | Qua | ProductName | Price |
CompanyD | 5/2/2012 | 500 | ABC | 10 |
CompanyB | 1/1/2012 | 1000 | DEF | 16 |
CompanyA | 9/10/2013 | 3069 | MNO | 100 |
至此,SQL 的逻辑执行过程处理完毕。
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