算法题目 Sql

Posted by SunQH Blog on February 13, 2021

Leetcode按类型

SQL

总结

SQL

MYSQL language

内联结:
select …
from … inner join
on …

外联结:
考虑到可能不是每个人都有地址信息,我们应该使用 outer join 而不是默认的 inner join select …
frome … left join …
on …

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# SQL
select FirstName, LastName, City, State
from Person left join Address
on Person.PersonID = Address.PersonID
;

🚩175. 组合两个表 - SQL架构

表1: Person

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+-------------+---------+
| 列名         | 类型     |
+-------------+---------+
| PersonId    | int     |
| FirstName   | varchar |
| LastName    | varchar |
+-------------+---------+
PersonId 是上表主键

表2: Address

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+-------------+---------+
| 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:

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FirstName, LastName, City, State

2️⃣

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select FirstName, LastName, City, State
from Person left join Address
on Person.PersonID = Address.PersonID
;