leetcode Database1（三）

一、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) | Date(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 |
+----+
分析：意思就是在Weather表中，写一个SQL查询与前一天相比温度更高的日期对应的ID。
代码：

# Write your MySQL query statement below
SELECT w1.Id 
FROM Weather w1 JOIN Weather w2 ON TO_DAYS(w1.Date)=TO_DAYS(w2.Date)+1 And w1.Temperature>w2.Temperature; 

　其中，TO_DAYS(date) 给定一个日期date, 返回一个天数 (从年份0开始的天数 )　

其他解法：

SELECT w1.Id FROM Weather w1, Weather w2 WHERE dateDiff(w1.Date,w2.Date) = 1 AND w1.Temperature > w2.Temperature;

　　其中，dateDiff() 函数返回两个日期之间的天数。

还有这样的方式：

 date_add(w1.date,interval 1 day)=w2.date

w2.Date = DATE_SUB(w1.Date, INTERVAL 1 DAY)

二、Delete Duplicate Emails

Write a SQL query to delete all duplicate email entries in a table named Person, keeping only unique emails based on its smallest Id.

+----+------------------+
| Id | Email            |
+----+------------------+
| 1  | john@example.com |
| 2  | bob@example.com  |
| 3  | john@example.com |
+----+------------------+
Id is the primary key column for this table.

For example, after running your query, the above Person table should have the following rows:

+----+------------------+
| Id | Email            |
+----+------------------+
| 1  | john@example.com |
| 2  | bob@example.com  |
+----+------------------+

 分析：意思就是删除Email列中重复项所在的行，而且保留的不重复行Id更小。

代码：

# Write your MySQL query statement below
DELETE p1
FROM Person p1, Person p2
WHERE p1.Email = p2.Email AND p1.Id > p2.Id

　其他解法：

DELETE FROM Person
    WHERE Id IN
    (SELECT P1.Id FROM Person AS P1, Person AS P2 
         WHERE P1.Id > P2.Id AND P1.Email = P2.Email);

报错：Runtime Error Message:You can't specify target table 'Person' for update in FROM clause

　所以得注意：In mysql you must't update a table while using select clause , You can only do that step by step . However ,you can use a middle table as :　

delete from Person where id not in( select t.id from ( select min(id) as id from Person group by email ) t )

　或：

MySQL Don't allow referring delete target table in sub query, a workaround is use ( select * from Person ) to get a new table.

delete from Person where Id in ( select p1.Id from (select * from Person) p1, (select * from Person) p2 where p1.Email = p2.Email and p1.Id > p2.Id )

另外ps：刚开始也想过用"SELECT DISTINCT Email from Person" ，但是注意到：Delete and Distinct are completely different, while delete alters the table, distinct only selects distinct values and doesn't alter table.　所以这样是不可行的！

三、Customers Who Never Order

Suppose that a website contains two tables, the Customers table and the Orders table. Write a SQL query to find all customers who never order anything.

Table: Customers.

+----+-------+
| Id | Name  |
+----+-------+
| 1  | Joe   |
| 2  | Henry |
| 3  | Sam   |
| 4  | Max   |
+----+-------+

Table: Orders.

+----+------------+
| Id | CustomerId |
+----+------------+
| 1  | 3          |
| 2  | 1          |
+----+------------+

Using the above tables as example, return the following:

+-----------+
| Customers |
+-----------+
| Henry     |
| Max       |
+-----------+
分析：题意为  假设一个网站包含两个表， 顾客表Customers和订单表Orders。编写一个SQL查询找出所有从未下过订单的顾客。
思路：使用NOT IN，NOT EXISTS，或者LEFT JOIN都是可以解决的。
解法一：
NOT IN

# Write your MySQL query statement below
SELECT Name
FROM Customers C WHERE C.Id not in (select O.CustomerId from Orders O);

　解法二：

NOT EXISTS

# Write your MySQL query statement below
SELECT Name FROM Customers c WHERE NOT EXISTS (SELECT CustomerId FROM Orders o WHERE o.CustomerId = c.id);

　解法三：

SELECT C.Name
FROM Customers  AS C LEFT OUTER JOIN Orders AS O
ON C.Id = O.CustomerId
WHERE O.CustomerId IS NULL;