Question

我有一个项目，我必须使用嵌套的选择和运算符，如EXISTS，IN，ALL，ANY，来查找拥有更多公寓的业主的名称。

我有2个表所有者和公寓

Owner
"ID" NUMBER(5,0), 
"NAME" VARCHAR2(20), 
"PHONE" NUMBER(10,0),
CONSTRAINT "PROPR" PRIMARY KEY ("ID")
USING INDEX  ENABLE; 

Apartment
"ID_AP" NUMBER(5,0), 
"ADDRESS" VARCHAR2(35), 
"SURFACE" NUMBER(10,0),
"ID" NUMBER(5,0),
CONSTRAINT "APART" PRIMARY KEY ("ID_AP")
USING INDEX  ENABLE;

为了找到有多个公寓的业主，我写了这个

SELECT name, id, count(id)
from apartment join owner
using (id)
group by id, name
having count(id)>1

但是如何使用嵌套的Selects和其中一个运算符EXISTS，IN，ALL，ANY？

Answer 1

例如：

                $http({
                    url: 'http://192.16.1.1:8080/restproj/v1/dealer/car',
                    method: "POST",
                    data: { 'message' : $scope.myMessage}
                })

Answer 2

我不太确定你写的查询是否正确。

您提供的正确查询将计算（id_ap）而不是count（id）

SELECT o.id, o.name, count(a.id_ap)
from apartment a
join owner o on o.id = a.id
group by o.id, o.name
having count(a.id_ap)>1

使用in子句的相同查询，但没有公寓数量，你只知道它是2或更多

select o.id, o.name
from owner o
where o.id in (select distinct(a.id) from apartment a
group by a.id having count(a.id_ap) > 1);

您在公寓表引用所有者表时也缺少外键约束。

Answer 3

WITH myView(name, id,appcount)
as
SELECT name, id, (select count(id) FROM apartment where id = ow.id) 
FROM owner ow

SELECT * from myView
WHERE appcount > 2

使用WITH创建内联视图并使用写入查询。（选择计数（id）FROM公寓，其中id = ow.id）在选择外部查询时将计算公寓数。

如何将查询更改为嵌套选择？

3 个答案: