最近在刷Leetcode中,遇到两道SQL小题,记录在此
join小例子:
181. Employees Earning More Than Their Managers
197. Rising Temperatureleft join 小例子:
175. Combine Two Tables
181. Employees Earning More Than Their Managers
The Employee
table holds all employees including their managers. Every employee has an Id, and there is also a column for the manager Id.
+----+-------+--------+-----------+ | Id | Name | Salary | ManagerId | +----+-------+--------+-----------+ | 1 | Joe | 70000 | 3 | | 2 | Henry | 80000 | 4 | | 3 | Sam | 60000 | NULL | | 4 | Max | 90000 | NULL | +----+-------+--------+-----------+
Given the Employee
table, write a SQL query that finds out employees who earn more than their managers. For the above table, Joe is the only employee who earns more than his manager.
+----------+ | Employee | +----------+ | Joe | +----------+
查询出比manager工资高的employee
select
e1.Name as Employee
from
Employee e1 join Employee e2
on e1.ManagerId = e2.Id
and e1.Salary > e2.Salary;
197. Rising Temperature
Given a Weather
table, write a SQL query to find all dates’ Ids with higher temperature compared to its previous (yesterday’s) dates.
+---------+------------------+------------------+ | Id(INT) | RecordDate(DATE) | Temperature(INT) | +---------+------------------+------------------+ | 1 | 2015-01-01 | 10 | | 2 | 2015-01-02 | 25 | | 3 | 2015-01-03 | 20 | | 4 | 2015-01-04 | 30 | +---------+------------------+------------------+
For example, return the following Ids for the above Weather
table:
+----+ | Id | +----+ | 2 | | 4 | +----+
查询出温度高于前一天的id,用到了DATEDIFF函数
# Write your MySQL query statement below
select
w1.Id from Weather w1 JOIN Weather w2
on DATEDIFF(w1.RecordDate, w2.RecordDate) = 1
and w1.Temperature > w2.Temperature;
175. Combine Two Tables
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
LEFT JOIN 关键字会从左表 (table_name1) 那里返回所有的行,即使在右表 (table_name2) 中没有匹配的行。
select
FirstName, LastName, City, State
from
Person left join Address
on Person.PersonId = Address.PersonId