oracle11g中的SQL语句

我有以下格式的表格：

ID      code   status
-----   -----  ------
1       Dept1    200
1       Dept2    500
1       Dept3    500
1       Dept3    200
2       Dept1    200
2       Dept2    500
2       Dept3    500
2       Dept3    500       
3       Dept1    200
3       Dept2    500
3       Dept3    500
3       Dept2    200
4       Dept1    500
4       Dept2    500
4       Dept3    500

我需要的输出是ID 1和ID 3

---

有谁能帮我写一个SQL来打印ID 1和ID 3，因为它们的状态是Dept 2和Dept 3都是200和500，Dept 1都是200

您需要

条件聚合

根据您的解释，只有

ID=3

满足所有条件

SELECT id 
FROM   Yourtable
GROUP  BY id 
HAVING Count(CASE WHEN "code" = 'Dept2' AND "status" = 200 THEN 1 END) = 1 
       AND Count(CASE WHEN "code" = 'Dept3' AND "status" = 500 THEN 1 END) = 1 
       AND Count(CASE WHEN "code" = 'Dept1' AND "status" = 200 THEN 1 END) = 1 

---

您的要求是查找ID与

代码

和

状态

的五种组合相匹配的记录。满足此要求的一个简单方法是将每个组合指定为单独的子查询，并使用INTERSECT set运算符将结果减少为唯一ID：

select id from your_table
where code = 'Dept1' and status = 200
intersect
select id from your_table
where code = 'Dept2' and status = 200
intersect
select id from your_table
where code = 'Dept2' and status = 500
intersect
select id from your_table
where code = 'Dept3' and status = 200
intersect
select id from your_table
where code = 'Dept3' and status = 500

---

1

没有

dept2和dept3

两个

200

和

500

。请再次解释条件，并提供输出示例？