SQL：单个人存在多个地址。如果存在则选择当前地址，否则选择永久地址

Posted 2023-02-16

【中文标题】SQL：单个人存在多个地址。如果存在则选择当前地址，否则选择永久地址

【英文标题】：SQL: Multiple addresses present for single person. Select current address if present otherwise select permanent address

【发布时间】：2021-03-16 13:06:26

【问题描述】：

我有这种情况，如果有人有当前地址，则需要选择该地址，否则选择永久地址。

我不想将Address 表与Person 加入两次，因为这会影响性能。

人

| person_id | name  | surname
+-----------+-------+----------
| 10        | ABC10 | XYZ10
| 11        | ABC11 | XYZ11
| 12        | ABC12 | XYZ12
| 13        | ABC13 | XYZ13

地址

| ID    | person_id | type      | address   | city
+-------+-----------+-----------+-----------+----------
| 1     | 10        | Permanent | addr 10P  | city 10P
| 2     | 10        | Current   | addr 10C  | city 10C
| 3     | 11        | Permanent | addr 11P  | city 11P
| 4     | 12        | Permanent | addr 12P  | city 12P
| 5     | 12        | Current   | addr 12C  | city 12C
| 6     | 13        | Permanent | addr 13P  | city 13P 

预期输出：

| person_id | name  | surname   | type      | address   | city
+-----------+-------+-----------+-----------+-----------+-----------
| 10        | ABC10 | XYZ10     | Current   | addr 10C  | city 10C
| 11        | ABC11 | XYZ11     | Permanent | addr 11P  | city 11P
| 12        | ABC12 | XYZ12     | Current   | addr 12C  | city 12C
| 13        | ABC13 | XYZ13     | Permanent | addr 13P  | city 13P

谢谢！

【参考方案1】：

一个选项使用横向连接和 fetch 子句：

select p.*, a.type, a.address, a.city
from person p
cross apply (
    select a.*
    from address a
    where a.person_id = p.person_id
    order by case a.type when 'Current' then 1 else 2 end
    fetch first row only
) a

您还可以使用窗口函数（这适用于所有 Oracle 版本）：

select p.*, a.type, a.address, a.city
from person p
inner join (
    select a.*, 
        row_number() over(partition by person_id order by case type when 'Current' then 1 else 2 end) as rn
    from address a
) a on a.person_id = p.person_id
where a.rn = 1

case 表达式可以轻松地使用更多可能的地址类型扩展优先级逻辑。如果您只有两个值，“Permanent”和“Current”，那么可以在两个查询中将其简化为：order by type（这会将“Current”放在首位）。

【参考方案2】：

您可以使用窗口函数来解决这个问题：

select p.*, a.type, a.address, a.city
from person p join
     (select a.*,
             row_number() over (partition by a.person_id order by a.type) as seqnum
      from address a
     ) a
     on a.person_id = p.person_id and a.seqnum = 1;

注意：这里假设类型只有“当前”和“永久”，所以它使用字符串排序来选择“当前”。

【参考方案3】：

您可以根据需要使用row_number或rank或dense_rank函数。

使用row_number的方法如下：

select * from
(select p.*, a.type, a.address, a.city,
        row_number() over (partition by a.person_id order by a.type) as rn
   from person p join address a
     on a.person_id = p.person_id
) where rn = 1;