一键选择Oracle SQL非唯一表别名 [英] Oracle SQL non-unique table alias in one Select

问题描述

有人知道为什么这两个表别名都为"x"吗?

Does somebody know why this works with both table alias "x"?

Select x.company_name  
      ,x.employee_name 
FROM company  x
JOIN employee x ON x.company_id = 5

我知道ID为5的JOIN毫无意义...

I know that the JOIN with id 5 makes no sense...

感谢您的教训！

推荐答案

下面的前两个查询是等效的.在联接的ON子句中，表别名x仅指使用该别名的最后一个表，因此仅限制employee表.

The first two queries below are equivalent. In the ON clause of the join the table alias x only refers to the last table to use that alias so only the employee table is restricted.

在SELECT和WHERE表达式中，x别名引用两个表-因此，在列名唯一的情况下可以成功引用它们，但在列名相同的情况下oracle引发异常(如果删除了注释，则在查询3中会发生这种情况.)

In the SELECT and WHERE expressions the x alias refers to both tables - so, where the column names are unique then they can be successfully referenced but where there are identical column names then oracle raises an ORA-00918: column ambiguously defined exception (as happens in query 3 if the comment is removed).

我找不到与此相关的任何文档，但看起来非常像个错误.

I can't find any documentation on this but it looks very like a bug.

SQL提琴

Oracle 11g R2架构设置:

CREATE TABLE employee (
  company_id    NUMBER(3),
  employee_name VARCHAR2(20)
);

CREATE TABLE company (
  company_id    NUMBER(3),
  company_name VARCHAR2(20)
);

INSERT INTO employee VALUES ( 4, 'Four Emp' );
INSERT INTO employee VALUES ( 5, 'Five Emp' );
INSERT INTO employee VALUES ( 6, 'Six Emp' );

INSERT INTO company VALUES ( 4, 'Four Company' );
INSERT INTO company VALUES ( 5, 'Five Company' );
INSERT INTO company VALUES ( 6, 'Six Company' );

查询1 :

SELECT *
FROM   company  x
       JOIN
       employee x
       ON x.company_id = 5

结果 :

Results:

| COMPANY_ID | COMPANY_NAME | EMPLOYEE_NAME |
|------------|--------------|---------------|
|          5 | Four Company |      Five Emp |
|          5 | Five Company |      Five Emp |
|          5 |  Six Company |      Five Emp |

查询2 :

SELECT *
FROM   company x
       CROSS JOIN
       (SELECT * FROM employee WHERE company_id = 5) x

结果 :

Results:

| COMPANY_ID | COMPANY_NAME | EMPLOYEE_NAME |
|------------|--------------|---------------|
|          5 | Four Company |      Five Emp |
|          5 | Five Company |      Five Emp |
|          5 |  Six Company |      Five Emp |

查询3 :

SELECT --x.company_id,
       x.company_name,
       x.employee_name
FROM   company x
       CROSS JOIN
       (SELECT * FROM employee WHERE company_id = 5) x

结果 :

Results:

| COMPANY_NAME | EMPLOYEE_NAME |
|--------------|---------------|
| Four Company |      Five Emp |
| Five Company |      Five Emp |
|  Six Company |      Five Emp |

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