Oracle声明不能正确计数

Question

我得到一个SQL语句 - 对我来说不能解释 - 奇怪的行为。 也许你会发现什么是错的：

当我使用

select count(*) from department 

我得到2755个结果

使用下面的语句

select 
     building1.street, building1.streetno, building1.plz, building1.city, dept1.buildingid 
    from 
     department dept1 
    left join 
     supporter sup 
    on 
     dept.supporterid = sup.id 
    left join 
     building building1 
    on 
     sup.buildingid = building1.ibuildingid 
    where 
     dept.usepostaladresssupporter = 1 
    union all 
    select 
     building2.street, building2.streetno, building2.plz, building2.city, dept2.buildingid 
    from 
     building building2 
    right join 
     tueks_department dept2 
    on 
     dept2.buildingid = building2.ibuildingid 
    where 
     dept2.usepostaladresssupporter = 0 

我得到了2755分的结果也声明。 但是，当我想结合这两种说法与左连接：

select count(*) from department 
    left join 
    (
     select 
      building1.street, building1.streetno, building1.plz, building1.city, dept1.buildingid 
     from 
      department dept1 
     left join 
      supporter sup 
     on 
      dept.supporterid = sup.id 
     left join 
      building building1 
     on 
      sup.buildingid = building1.ibuildingid 
     where 
      dept.usepostaladresssupporter = 1 
     union all 
     select 
      building2.street, building2.streetno, building2.plz, building2.city, dept2.buildingid 
     from 
      building building2 
     right join 
      tueks_department dept2 
     on 
      dept2.buildingid = building2.ibuildingid 
     where 
      dept2.usepostaladresssupporter = 0 
    ) postadress 
    on 
    department.buildingid = postadress.buildingid; 

我得到了3648513个结果。

我的期望是，我只得到2755结果。 错误在哪里？

感谢您的帮助！

Answer 1

我认为buildingid不是唯一（我的推论是成立的，它不能是唯一的）

想象一下下面这个简单的表

表A

create TableA (name VARCHAR(32)); 
insert into TableA values ('Lieven'); 
insert into TableA values ('Lieven');  

表B

create TableB (name VARCHAR(32)); 
insert into TableB values ('Lieven'); 
insert into TableB values ('Lieven'); 
insert into TableB values ('AnyOtherValue'); 

Select语句

select * from TableA a left outer join TableB b on a.name = b.name 

由于的TableA每个记录与匹配的TableB每个记录，其中name是平等的，这将导致4条（该AnyOtherValue被dissmissed因为它不匹配）

• 的TableA的第一条记录与二三记录`表B返回的“
• 的TableA第二个记录与二三记录`表B返回的”

Answer 2

查询

select 
     building1.street, building1.streetno, building1.plz, building1.city, dept1.buildingid 
    from 
     department dept1 
    left join 
     supporter sup 
    on 
     dept.supporterid = sup.id 
    left join 
     building building1 
    on 
     sup.buildingid = building1.ibuildingid 
    where 
     dept.usepostaladresssupporter = 1 
    union all 
    select 
     building2.street, building2.streetno, building2.plz, building2.city, dept2.buildingid 
    from 
     building building2 
    right join 
     tueks_department dept2 
    on 
     dept2.buildingid = building2.ibuildingid 
    where 
     dept2.usepostaladresssupporter = 0 

将返回一个行，每个部门有usepostaladresssupporter作为0或1（注意与其他值的记录不会被包括，这可能会或可能不会成为一个问题取决于此列的限制）。

此查询结果的唯一键可能类似于departmentid（您需要在选择标准中包含该列）。

所以正确的查询应该是这个样子：

select * from department 
    left join 
    (
     select 
      building1.street, building1.streetno, building1.plz, building1.city, dept1.departmentid 
     from 
      department dept1 
     left join 
      supporter sup 
     on 
      dept.supporterid = sup.id 
     left join 
      building building1 
     on 
      sup.buildingid = building1.ibuildingid 
     where 
      dept.usepostaladresssupporter = 1 
     union all 
     select 
      building2.street, building2.streetno, building2.plz, building2.city, dept2.departmentid 
     from 
      building building2 
     right join 
      tueks_department dept2 
     on 
      dept2.buildingid = building2.ibuildingid 
     where 
      dept2.usepostaladresssupporter = 0 
    ) postadress 
    on 
    department.departmentid = postadress.departmentid; 

您的查询就会出错的数据是这样的：

DepartmentID的BuildingId名称

11 DEPT1

2 2部门2

3 2 Dept3

的倍增效应并不完全等于deptcount * deptcount，而是它是buildingcount * buildingcount + deptcount - buildingcount