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;

  

 
 

 

 

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