PHP算法练习2:(175. 组合两个表)

练习地址:https://leetcode-cn.com/problems/combine-two-tables/

表1: Person

+-------------+---------+
| 列名 | 类型 |
+-------------+---------+
| PersonId | int |
| FirstName | varchar |
| LastName | varchar |
+-------------+---------+
PersonId 是上表主键
表2: Address

+-------------+---------+
| 列名 | 类型 |
+-------------+---------+
| AddressId | int |
| PersonId | int |
| City | varchar |
| State | varchar |
+-------------+---------+
AddressId 是上表主键

编写一个 SQL 查询，满足条件：无论 person 是否有地址信息，都需要基于上述两表提供 person 的以下信息： 

FirstName, LastName, City, State

答案:

select
 a.FirstName,a.LastName,b.City,b.State
 from
 Person  a left join Address b
 on
  a.PersonId =b.PersonId

原文地址：https://www.cnblogs.com/8013-cmf/p/12411061.html