0

我有一个关于外连接的新手问题。我有 5 个具有以下结构的表。

employee (employee_id, name, address)
employee_benefits (employee_id, benefit_id, join_date)
insurance (insurance_id, name)
insurance_benefits (insurance_id, benefit_id)
benefit (benefit_id, title)

对于选定的员工，我需要选择所有保险，其福利与员工可获得的全部或部分福利相匹配。例子：

benefit(
    bn1, 1st benefit;
    bn2, 2nd benefit;
    bn3, 3rd benefit;
    bn4, 4th benefit; 
)

employee(
    123, bill jones, 123 main st;
    321, alex baldwin, 222 state st;
)

employee_benefits(
    123, bn1;
    123, bn2;
    123, bn4; 
    321, bn3;
    321, bn4;
)

insurance(
    ins1, 1st insurance;
    ins2, 2nd insurance;
    ins3, 3rd insurance
)

insurance_benefits(
    ins1, bn1;
    ins1, bn2;
    ins2, bn1;
    ins2, bn3;
    ins2, bn4;
    ins3, bn2;
    ins3, bn4;
)

对于员工 123。我需要选择具有完整福利列表或员工持有的福利子集但不包括员工未持有福利的情况的所有保险。在上面的示例中，它将是 ins1 和 ins3。

有没有办法使用外连接来完成上述工作？

4

1 回答 1

0

您可以使用外部连接来执行此操作，但我认为将 - 子句与子查询一起使用更直观IN。

员工未持有的福利123：

SELECT benefit_id
  FROM benefit
 WHERE benefit_id NOT IN
        ( SELECT benefit_id
            FROM employee_benefits
           WHERE employee_id = 123
        )
;

提供上述任何福利的保险计划：

SELECT DISTINCT insurance_id
  FROM insurance_benefits
 WHERE benefit_id NOT IN
        ( SELECT benefit_id
            FROM employee_benefits
           WHERE employee_id = 123
        )
;

不提供上述任何福利的保险计划：

SELECT insurance_id
  FROM insurance
 WHERE insurance_id NOT IN
        ( SELECT DISTINCT insurance_id
            FROM insurance_benefits
           WHERE benefit_id NOT IN
                  ( SELECT benefit_id
                      FROM employee_benefits
                     WHERE employee_id = 123
                  )
        )
;
于 2013-07-01T00:44:20.630 回答