175. Combine Two Tables
https://leetcode.com/problems/combine-two-tables/description/
Description
Person 表：
+-------------+---------+
| Column Name | Type |
+-------------+---------+
| PersonId | int |
| FirstName | varchar |
| LastName | varchar |
+-------------+---------+
PersonId is the primary key column for this table.
Address 表：
+-------------+---------+
| Column Name | Type |
+-------------+---------+
| AddressId | int |
| PersonId | int |
| City | varchar |
| State | varchar |
+-------------+---------+
AddressId is the primary key column for this table.
查找 FirstName, LastName, City, State 數(shù)據(jù)吐辙，而不管一個(gè)用戶有沒有填地址信息键耕。
SQL Schema
DROP TABLE Person;
CREATE TABLE Person ( PersonId INT, FirstName VARCHAR ( 255 ), LastName VARCHAR ( 255 ) );
DROP TABLE Address;
CREATE TABLE Address ( AddressId INT, PersonId INT, City VARCHAR ( 255 ), State VARCHAR ( 255 ) );
INSERT INTO Person ( PersonId, LastName, FirstName )
VALUES
( 1, 'Wang', 'Allen' );
INSERT INTO Address ( AddressId, PersonId, City, State )
VALUES
( 1, 2, 'New York City', 'New York' );
Solution
使用左外連接谷誓。
SELECT FirstName, LastName, City, State
FROM Person AS P LEFT JOIN Address AS A
ON P.PersonId = A.PersonId;
181. Employees Earning More Than Their Managers
https://leetcode.com/problems/employees-earning-more-than-their-managers/description/
Description
Employee 表：
+----+-------+--------+-----------+
| Id | Name | Salary | ManagerId |
+----+-------+--------+-----------+
| 1 | Joe | 70000 | 3 |
| 2 | Henry | 80000 | 4 |
| 3 | Sam | 60000 | NULL |
| 4 | Max | 90000 | NULL |
+----+-------+--------+-----------+
查找所有員工，它們的薪資大于其經(jīng)理薪資橘券。
SQL Schema
CREATE TABLE Employee ( Id INT, NAME VARCHAR ( 255 ), Salary INT, ManagerId INT );
INSERT INTO Employee ( Id, NAME, Salary, ManagerId )
VALUES
( '1', 'Joe', '70000', '3' ),
( '2', 'Henry', '80000', '4' ),
( '3', 'Sam', '60000', NULL ),
( '4', 'Max', '90000', NULL );
Solution
SELECT E1.Name
FROM Employee AS E1, Employee AS E2
WHERE E1.ManagerId = E2.Id AND E1.Salary > E2.Salary;
183. Customers Who Never Order
https://leetcode.com/problems/customers-who-never-order/description/
Description
Curstomers 表：
+----+-------+
| Id | Name |
+----+-------+
| 1 | Joe |
| 2 | Henry |
| 3 | Sam |
| 4 | Max |
+----+-------+
Orders 表：
+----+------------+
| Id | CustomerId |
+----+------------+
| 1 | 3 |
| 2 | 1 |
+----+------------+
查找沒有訂單的顧客信息：
+-----------+
| Customers |
+-----------+
| Henry |
| Max |
+-----------+
SQL Schema
DROP TABLE Customers;
CREATE TABLE Customers ( Id INT, NAME VARCHAR ( 255 ) );
DROP TABLE Orders;
CREATE TABLE Orders ( Id INT, CustomerId INT );
INSERT INTO Customers ( Id, NAME )
VALUES
( '1', 'Joe' ),
( '2', 'Henry' ),
( '3', 'Sam' ),
( '4', 'Max' );
INSERT INTO Orders ( Id, CustomerId )
VALUES
( '1', '3' ),
( '2', '1' );
Solution
左外鏈接
SELECT C.Name AS Customers
FROM Customers AS C LEFT JOIN Orders AS O
ON C.Id = O.CustomerId
WHERE O.CustomerId IS NULL;
子查詢
SELECT C.Name AS Customers
FROM Customers AS C
WHERE C.Id NOT IN (
SELECT CustomerId
FROM Orders
);
184. Department Highest Salary
https://leetcode.com/problems/department-highest-salary/description/
Description
Employee 表：
+----+-------+--------+--------------+
| Id | Name | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1 | Joe | 70000 | 1 |
| 2 | Henry | 80000 | 2 |
| 3 | Sam | 60000 | 2 |
| 4 | Max | 90000 | 1 |
+----+-------+--------+--------------+
Department 表：
+----+----------+
| Id | Name |
+----+----------+
| 1 | IT |
| 2 | Sales |
+----+----------+
查找一個(gè) Department 中收入最高者的信息：
+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT | Max | 90000 |
| Sales | Henry | 80000 |
+------------+----------+--------+
SQL Schema
DROP TABLE Employee;
CREATE TABLE Employee ( Id INT, NAME VARCHAR ( 255 ), Salary INT, DepartmentId INT );
DROP TABLE Department;
CREATE TABLE Department ( Id INT, NAME VARCHAR ( 255 ) );
INSERT INTO Employee ( Id, NAME, Salary, DepartmentId )
VALUES
( 1, 'Joe', 70000, 1 ),
( 2, 'Henry', 80000, 2 ),
( 3, 'Sam', 60000, 2 ),
( 4, 'Max', 90000, 1 );
INSERT INTO Department ( Id, NAME )
VALUES
( 1, 'IT' ),
( 2, 'Sales' );
Solution
SELECT D.Name AS Department, E.Name AS Employee, E.Salary
FROM Employee AS E, Department AS D,
(SELECT DepartmentId, MAX(Salary) AS Salary
FROM Employee
GROUP BY DepartmentId) AS M
WHERE E.DepartmentId = D.Id
AND E.DepartmentId = M.DepartmentId
AND E.Salary = M.Salary;
