LeetCode #175 CombineTwoTables. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,그래프,Graph,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접, 데이터베이스, sql, query, 쿼리
Runtime: 599 ms, faster than 5.61% of MySQL online submissions for Combine Two Tables.
Memory Usage: 0B, less than 100.00% of MySQL online submissions for Combine Two Tables.
전체 조인을 해서 조회하면 내용이 비어있어서 이상했는데, Left 조인을 사용해야함
왜냐면 새로 테이블 만드는게 아니라 Person 테이블에서 조회해야함
The question said regardless someone has address or not, which means the address might be empty. In such case if you use full outer join, all information will be combined together, and if you use inner join, the empty value will be ignored. So left join is the best choice.
LeetCode #175
Q.
Table:
Person
+-------------+---------+
| Column Name | Type |
+-------------+---------+
| PersonId | int |
| FirstName | varchar |
| LastName | varchar |
+-------------+---------+
PersonId is the primary key column for this table.
Table:
Address
+-------------+---------+
| Column Name | Type |
+-------------+---------+
| AddressId | int |
| PersonId | int |
| City | varchar |
| State | varchar |
+-------------+---------+
AddressId is the primary key column for this table.
Write a SQL query for a report that provides the following information for each person in the Person table, regardless if there is an address for each of those people:
FirstName, LastName, City, State
Process
// Process
//1. Select in Person / Address Tables
//2. Columns FirstName, LastName / City, State in each columns
//3. When selecting, join to Person table's side
//4. Join on each PersonId column values are the same
// 처리과정
//1. Person 테이블과, Address 테이블에서, 조회를 하는데
//2. 각각 FirstName, LastName / City, State 를 조회한다.
//3. 두 테이블을 Person 테이블 쪽으로 Join 해서 조회하는데,
//4. PersonId 가 같은 것을 기준으로 조인한다.
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
# Write your MySQL query statement below
SELECT Person.FirstName, Person.LastName, Address.City, Address.State
FROM Person LEFT JOIN Address
ON Person.PersonId = Address.PersonId;
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