# 175. 组合两个表

### Link

{% embed url="<https://leetcode.cn/problems/combine-two-tables/>" %}

### Problem

编写一个SQL查询来报告 Person 表中每个人的姓、名、城市和州。如果 personId 的地址不在 Address 表中，则报告为空  null 。

以 任意顺序 返回结果表。

查询结果格式如下所示。

示例 1：

{% code overflow="wrap" %}

```
输入: 
Person表:
+----------+----------+-----------+
| personId | lastName | firstName |
+----------+----------+-----------+
| 1        | Wang     | Allen     |
| 2        | Alice    | Bob       |
+----------+----------+-----------+
Address表:
+-----------+----------+---------------+------------+
| addressId | personId | city          | state      |
+-----------+----------+---------------+------------+
| 1         | 2        | New York City | New York   |
| 2         | 3        | Leetcode      | California |
+-----------+----------+---------------+------------+
输出: 
+-----------+----------+---------------+----------+
| firstName | lastName | city          | state    |
+-----------+----------+---------------+----------+
| Allen     | Wang     | Null          | Null     |
| Bob       | Alice    | New York City | New York |
+-----------+----------+---------------+----------+
解释: 
地址表中没有 personId = 1 的地址，所以它们的城市和州返回 null。
addressId = 1 包含了 personId = 2 的地址信息
```

{% endcode %}

### Solution

{% tabs %}
{% tab title="MySQL" %}

```sql
# Write your MySQL query statement below
select firstName, lastName, city, state 
from Person left join Address on Person.personId = Address.personId
```

{% endtab %}
{% endtabs %}