19.?Students With Invalid Departments
Table:?Departments
+---------------+---------+
| Column Name? | Type? ? |
+---------------+---------+
| id? ? ? ? ? ? | int? ? |
| name? ? ? ? ? | varchar |
+---------------+---------+
id is the primary key of this table.
The table has information about the id of each department of a university.
Table:?Students
+---------------+---------+
| Column Name? | Type? ? |
+---------------+---------+
| id? ? ? ? ? ? | int? ? |
| name? ? ? ? ? | varchar |
| department_id | int? ? |
+---------------+---------+
id is the primary key of this table.
The table has information about the id of each student at a university and the id of the department he/she studies at.
Write an SQL query to find the id and the name of all students who are enrolled in departments that no longer exists. Return the result table in any order.
SELECT a.id, a.name
FROM students AS a
LEFT JOIN departments AS b
ON a.department_id = b.id
WHERE b.id IS NULL
;
20.?Average Selling Price
Table:?Prices
+---------------+---------+
| Column Name? | Type? ? |
+---------------+---------+
| product_id? ? | int? ? |
| start_date? ? | date? ? |
| end_date? ? ? | date? ? |
| price? ? ? ? | int? ? |
+---------------+---------+
(product_id, start_date, end_date) is the primary key for this table.
Each row of this table indicates the price of the product_id in the period from start_date to end_date.
For each product_id there will be no two overlapping periods. That means there will be no two intersecting periods for the same product_id.
Table:?UnitsSold
+---------------+---------+
| Column Name? | Type? ? |
+---------------+---------+
| product_id? ? | int? ? |
| purchase_date | date? ? |
| units? ? ? ? | int? ? |
+---------------+---------+
There is no primary key for this table, it may contain duplicates.
Each row of this table indicates the date, units and product_id of each product sold.
Write an SQL query to find the average selling price for each product.
average_price?should be?rounded to 2 decimal places.
SELECT a.product_id,
? ? ? ? ? ? ? ? ROUND(IFNULL(SUM(a.units*b.price)/SUM(a.units), 0), 2) AS average_price
FROM unitssold AS a
LEFT JOIN prices AS b
ON a.product_id = b.product_id AND a.purchase_date BETWEEN b.start_date AND end_date
GROUP BY a.product_id
;
21.?Actors and Directors Who Cooperated At Least Three Times
Table:?ActorDirector
+-------------+---------+
| Column Name | Type? ? |
+-------------+---------+
| actor_id? ? | int? ? |
| director_id | int? ? |
| timestamp? | int? ? |
+-------------+---------+
timestamp is the primary key column for this table.
Write a SQL query for a report that provides the pairs?(actor_id, director_id)?where the actor have cooperated with the director at least 3 times.
SELECT?actor_id, director_id
FROM
(SELECT actor_id, director_id, COUNT(timestamp) AS num
FROM actordirector
GROUP BY actor_id, director_id) AS a
WHERE num >= 3
;
22.?Customer Placing the Largest Number of Orders
Query the?customer_number?from the?orders?table for the customer who has placed the largest number of orders.
It is guaranteed that exactly one customer will have placed more orders than any other customer.
The?orders?table is defined as follows:
| Column? ? ? ? ? ? | Type? ? ? |
|-------------------|-----------|
| order_number (PK) | int? ? ? |
| customer_number? | int? ? ? |
| order_date? ? ? ? | date? ? ? |
| required_date? ? | date? ? ? |
| shipped_date? ? ? | date? ? ? |
| status? ? ? ? ? ? | char(15)? |
| comment? ? ? ? ? | char(200) |
SELECT?customer_number
FROM
(SELECT?customer_number, COUNT(order_number) AS num
FROM orders
GROUP BY customer_number
ORDER BY num desc
LIMIT 1) AS a
;
如果這道題不能保證最后的結果只有一個customer呢？
solution1:
SELECT customer_number
FROM
(SELECT customer_number, DENSE_RANK() OVER(ORDER BY num DESC) AS rk
FROM
(SELECT customer_number, COUNT(order_number) AS num
FROM orders
GROUP BY customer_number) AS a ) AS a
WHERE rk = 1
;
solution 2:
SELECT customer_number
FROM
(SELECT?customer_number, COUNT(order_number) AS num
FROM orders
GROUP BY customer_number) AS a
WHERE num = (SELECT MAX(num) FROM (SELECT?customer_number, COUNT(order_number) AS num
FROM orders
GROUP BY customer_number
) AS a)
;
一般這種情況適合用dense_rank，不適合用having去進行把聚合函數放入子查詢。
23.?Product Sales Analysis I
Table:Sales
+-------------+-------+
| Column Name | Type? |
+-------------+-------+
| sale_id? ? | int? |
| product_id? | int? |
| year? ? ? ? | int? |
| quantity? ? | int? |
| price? ? ? | int? |
+-------------+-------+
(sale_id, year) is the primary key of this table.product_id is a foreign key toProducttable.Note that the price is per unit.
Table:Product
+--------------+---------+
| Column Name? | Type? ? |
+--------------+---------+
| product_id? | int? ? |
| product_name | varchar |
+--------------+---------+
product_id is the primary key of this table。
Write an SQL query that reports all?product names?of the products in the?Sales?table along with their selling?year?and?price.
SELECT product_name, year, price
FROM sales AS a
LEFT JOIN product AS b
ON a.product_id = b.product_id
;
此題的result當中的price不是total price， 直接選就可以。
24.?Shortest Distance in a Line
Table?point?holds the x coordinate of some points on x-axis in a plane, which are all integers.
Write a query to find the shortest distance between two points in these points.
| x? |
|-----|
| -1? |
| 0? |
| 2? |
Note:?Every point is unique, which means there is no duplicates in table?point.
SELECT MIN(distance) AS shortest
FROM
(SELECT ABS(a.x-b.x) AS distance
FROM point AS a, point AS b
HAVING distance <> 0) AS a
;
SELECT MIN(ABS(a.x-b.x)) AS shortest
FROM point AS a, point AS b
WHERE?ABS(a.x-b.x) <> 0
;
這個方案中，不能用having shortest <> 0，因為shortest已經選出了最小值，再having最小值不等于0，是無效的。
SELECT MIN(ABS(a.x-b.x)) AS shortest
FROM point AS a
INNER JOIN point AS b
ON a.x <> b.x
;
25.??Sales Analysis III
Table:Product
+--------------+---------+
| Column Name? | Type? ? |
+--------------+---------+
| product_id? | int? ? |
| product_name | varchar |
| unit_price? | int? ? |
+--------------+---------+
product_id is the primary key of this table.
Table:Sales
+-------------+---------+
| Column Name | Type? ? |
+-------------+---------+
| seller_id? | int? ? |
| product_id? | int? ? |
| buyer_id? ? | int? ? |
| sale_date? | date? ? |
| quantity? ? | int? ? |
| price? ? ? | int? ? |
+------ ------+---------+
This table has no primary key, it can have repeated rows.
product_id is a foreign key to Product table.
Write an SQL query that reports the?products?that were?only?sold in spring 2019. That is, between 2019-01-01?and?2019-03-31?inclusive.
SELECT a.product_id, b.product_name
FROM
(SELECT DISTINCT product_id
FROM sales
WHERE product_id NOT IN (SELECT product_id FROM sales WHERE sale_date NOT BETWEEN '2019-01-01' AND '2019-04-01')) AS a
LEFT JOIN product AS b
ON a.product_id = b.product_id
;
這個題是一個很大的陷阱，如果題中出現了讓你選擇某一個column，這個column只能出現在一個類別里，不能出現在另一個類別里，那么這個時候就要小心，不能直接選它在一個類別里，而是要讓它不在另一個類別里，剩下的自然就是在你想選的那個類別里了。還有就是，distinct最好放在join之前。
此題還可以跟sales analysis II做類比去做。
26.?Biggest Single Number
Table?my_numbers?contains many numbers in column?num?including duplicated ones.
Can you write a SQL query to find the biggest number, which only appears once.
+---+
|num|
+---+
| 8 |
| 8 |
| 3 |
| 3 |
| 1 |
| 4 |
| 5 |
| 6 |
If there is no such number, just output?null.
SELECT
IFNULL((SELECT num
FROM
(SELECT num, COUNT(num) AS n
FROM my_numbers
GROUP BY num
HAVING n = 1) AS a
ORDER BY num DESC
LIMIT 1), NULL) AS num
;
SELECT
(SELECT num
FROM
(SELECT num, COUNT(num) AS n
FROM my_numbers
GROUP BY num
HAVING n = 1) AS a
ORDER BY num DESC
LIMIT 1) AS num
;
如果沒有符合條件的話，正常返回是什么也沒有。但是如果想表達“如果什么都沒有則返回NULL”，那么有兩種方式：一種是加IF NULL然后SELECT， 一種是只加SELECT，這兩種都可以。
27.?Big Countries
There is a table?World
+-----------------+------------+------------+--------------+---------------+
| name? ? ? ? ? ? | continent? | area? ? ? | population? | gdp? ? ? ? ? |
+-----------------+------------+------------+--------------+---------------+
| Afghanistan? ? | Asia? ? ? | 652230? ? | 25500100? ? | 20343000? ? ? |
| Albania? ? ? ? | Europe? ? | 28748? ? ? | 2831741? ? ? | 12960000? ? ? |
| Algeria? ? ? ? | Africa? ? | 2381741? ? | 37100000? ? | 188681000? ? |
| Andorra? ? ? ? | Europe? ? | 468? ? ? ? | 78115? ? ? ? | 3712000? ? ? |
| Angola? ? ? ? ? | Africa? ? | 1246700? ? | 20609294? ? | 100990000? ? |
+-----------------+------------+------------+--------------+---------------+
A country is big if it has an area of bigger than 3 million square km or a population of more than 25 million.
Write a SQL solution to output big countries' name, population and area.
SELECT name, area, population
FROM world
WHERE area >= 3000000 OR population >= 25000000
;
28.?Group Sold Products By The Date
Table?Activities:
+-------------+---------+
| Column Name | Type? ? |
+-------------+---------+
| sell_date? | date? ? |
| product? ? | varchar |
+-------------+---------+
There is no primary key for this table, it may contains duplicates.
Each row of this table contains the product name and the date it was sold in a market.
Write an SQL query to find for each date, the number of distinct products sold and their names.
The sold-products names for each date should be sorted lexicographically.?
Return the result table ordered by?sell_date.
SELECT sell_date, COUNT(DISTINCT product) AS num_sold, GROUP_CONCAT(DISTINCT product) AS products
FROM
(SELECT *
FROM activities
ORDER BY product) AS a
GROUP BY sell_date
;
當要把group_concat里面的內容排序時，我們應該先把raw data進行order，然后再group_concat，它自然就排好了。另外，group_concat不支持window function，至少在mysql是這樣。
29.?Find Customer Referee
Given a table?customer?holding customers information and the referee.
+------+------+-----------+
| id? | name | referee_id|
+------+------+-----------+
|? ? 1 | Will |? ? ? NULL |
|? ? 2 | Jane |? ? ? NULL |
|? ? 3 | Alex |? ? ? ? 2 |
|? ? 4 | Bill |? ? ? NULL |
|? ? 5 | Zack |? ? ? ? 1 |
|? ? 6 | Mark |? ? ? ? 2 |
+------+------+-----------+
Write a query to return the list of customers?NOT?referred by the person with id '2'.
SELECT DISTINCT name
FROM customer
WHERE referee_id <> 2 OR referee_id IS NULL
;
30.?Game Play Analysis II
Table:Activity
+--------------+---------+
| Column Name? | Type? ? |
+--------------+---------+
| player_id? ? | int? ? |
| device_id? ? | int? ? |
| event_date? | date? ? |
| games_played | int? ? |
+--------------+---------+
(player_id, event_date) is the primary key of this table.
This table shows the activity of players of some game.
Each row is a record of a player who logged in and played a number of games (possibly 0) before logging out on some day using some device.
Write a?SQL query that reports?the?device?that is first logged in?for each player.
SELECT a.player_id, b.device_id
FROM
(SELECT player_id, MIN(event_date) AS first_login_date
FROM activity
GROUP BY player_id) AS a
INNER JOIN activity AS b
ON a.player_id = b.player_id AND a.first_login_date = b.event_date
;
31.?Replace Employee ID With The Unique Identifier
Table:?Employees
+---------------+---------+
| Column Name? | Type? ? |
+---------------+---------+
| id? ? ? ? ? ? | int? ? |
| name? ? ? ? ? | varchar |
+---------------+---------+
id is the primary key for this table.
Each row of this table contains the id and the name of an employee in a company.
Table:?EmployeeUNI
+---------------+---------+
| Column Name? | Type? ? |
+---------------+---------+
| id? ? ? ? ? ? | int? ? |
| unique_id? ? | int? ? |
+---------------+---------+
(id, unique_id) is the primary key for this table.
Each row of this table contains the id and the corresponding unique id of an employee in the company.
Write an SQL query to?show?the?unique ID?of each user, If a user doesn't have a unique ID replace just show null. Return the result table in?any?order.
SELECT IFNULL(DISTINCT b.unique_id, NULL) AS unique_id, name
FROM employees AS a
LEFT JOIN employeeuni AS b
ON a.id = b.id
;
32.?Not Boring Movies
X city opened a new cinema, many people would like to go to this cinema. The cinema also gives out a poster indicating the movies’ ratings and descriptions.
Please write a SQL query to output movies with an odd numbered ID and a description that is not 'boring'. Order the result by rating.
For example, table?cinema:
+---------+-----------+--------------+-----------+
|? id? ? | movie? ? |? description |? rating? |
+---------+-----------+--------------+-----------+
|? 1? ? | War? ? ? |? great 3D? |? 8.9? ? |
|? 2? ? | Science? |? fiction? ? |? 8.5? ? |
|? 3? ? | irish? ? |? boring? ? |? 6.2? ? |
|? 4? ? | Ice song? |? Fantacy? ? |? 8.6? ? |
|? 5? ? | House card|? Interesting|? 9.1? ? |
+---------+-----------+--------------+-----------+
SELECT id, movie, description, rating
FROM cinema
WHERE MOD(id, 2) = 1 AND description NOT LIKE 'boring'
ORDER BY rating DESC
;
33.?Classes More Than 5 Students
There is a table?courses?with columns:?student?and?class
Please list out all classes which have more than or equal to 5 students.
Note:?The students should not be counted duplicate in each course.
SELECT class
FROM
(SELECT class, COUNT(DISTINCT student) AS num
FROM courses
GROUP BY class
HAVING num >= 5) AS a
;
34.?Fix Product Name Format (跳過，不考該題型)
Table:Sales
+--------------+---------+
| Column Name? | Type? ? |
+--------------+---------+
| sale_id? ? ? | int? ? |
| product_name | varchar |
| sale_date? ? | date? ? |
+--------------+---------+
sale_id is the primary key for this table.
Since table Sales was filled manually in the year 2000, product_name may contain leading and/or trailing white spaces, also they are case-insensitive.
Write an SQL query to report:
product_name in lowercase without leading or trailing white spaces,
sale_date in the format('YYYY-MM'),
total the number of times the product was sold in this month.
Return the result table ordered by product_name in?ascending order, in case of a tie order it?by sale_date in?ascending order.
35.?Patients With a Condition
Table:Patients
+--------------+---------+
| Column Name? | Type? ? |
+--------------+---------+
| patient_id? | int? ? |
| patient_name | varchar |
| conditions? | varchar |
+--------------+---------+
patient_id is the primary key for this table.
'conditions' contains 0 or more code separated by spaces.
This table contains information of the patients in the hospital.
Write an SQL query to report the patient_id, patient_name all conditions of patients who have Type I Diabetes. Type I Diabetes always starts with?DIAB1?prefix.
SELECT?patient_id, patient_name, conditions
FROM patients
WHERE conditions LIKE '%DIAB1%'
;
36.?Find Users With Valid E-Mails (不考這個題型，跳過)
Table:?Users
+---------------+---------+
| Column Name? | Type? ? |
+---------------+---------+
| user_id? ? ? | int? ? |
| name? ? ? ? ? | varchar |
| mail? ? ? ? ? | varchar |
+---------------+---------+
user_id is the primary key for this table.
This table contains information of the users signed up in a website. Some e-mails are invalid.
Write an SQL query to?find the?users who have?valid emails.
A valid e-mail has a prefix name and a domain where:?
The prefix name?is a string that may contain letters (upper or lower case), digits, underscore?'_', period?'.'?and/or dash?'-'. The prefix name?must?start with a letter.
The domain?is?'@leetcode.com'.
37.?Customer Order Frequency
Table:?Customers
+---------------+---------+
| Column Name? | Type? ? |
+---------------+---------+
| customer_id? | int? ? |
| name? ? ? ? ? | varchar |
| country? ? ? | varchar |
+---------------+---------+
customer_id is the primary key for this table.
This table contains information of the customers in the company.
Table:?Product
+---------------+---------+
| Column Name? | Type? ? |
+---------------+---------+
| product_id? ? | int? ? |
| description? | varchar |
| price? ? ? ? | int? ? |
+---------------+---------+
product_id is the primary key for this table.
This table contains information of the products in the company.
price is the product cost.
Table:?Orders
+---------------+---------+
| Column Name? | Type? ? |
+---------------+---------+
| order_id? ? ? | int? ? |
| customer_id? | int? ? |
| product_id? ? | int? ? |
| order_date? ? | date? ? |
| quantity? ? ? | int? ? |
+---------------+---------+
order_id is the primary key for this table.
This table contains information on customer orders.
customer_id is the id of the customer who bought "quantity" products with id "product_id".
Order_date is the date in format ('YYYY-MM-DD') when the order was shipped.
Write an SQL query to?report the?customer_id and customer_name of customers who have spent at least $100 in each month of June and July 2020.
WITH temp_1 AS (SELECT customer_id, MONTH(order_date) AS mon, SUM(o.quantity*p.price) AS price
FROM orders AS o
LEFT JOIN product AS p
ON o.product_id = p.product_id
WHERE YEAR(o.order_date) = 2020 and MONTH(order_date) IN (6,7)
GROUP BY customer_id, MONTH(order_date)
HAVING price >= 100),
temp_2 AS (SELECT DISTINCT customer_id FROM temp_1 WHERE customer_id IN (SELECT customer_id FROM (SELECT customer_id, COUNT(mon) FROM temp_1 GROUP BY customer_id HAVING COUNT(mon) = 2) AS a))
SELECT c.customer_id, c.name FROM temp_2 LEFT JOIN customers AS c ON temp_2.customer_id = c.customer_id
;
這道題的邏輯其實并不復雜，但是涉及到的子查詢會比較多，容易搞錯，要很細心。耗時也是蠻長的。
38.?Friendly Movies Streamed Last Month
Table:?TVProgram
+---------------+---------+
| Column Name? | Type? ? |
+---------------+---------+
| program_date? | date? ? |
| content_id? ? | int? ? |
| channel? ? ? | varchar |
+---------------+---------+
(program_date, content_id) is the primary key for this table.
This table contains information of the programs on the TV.
content_id is the id of the program in some channel on the TV.
Table:?Content
+------------------+---------+
| Column Name? ? ? | Type? ? |
+------------------+---------+
| content_id? ? ? | varchar |
| title? ? ? ? ? ? | varchar |
| Kids_content? ? | enum? ? |
| content_type? ? | varchar |
+------------------+---------+
content_id is the primary key for this table.
Kids_content is an enum that takes one of the values ('Y', 'N') where:
'Y' means is content for kids otherwise 'N' is not content for kids.
content_type?is the category of the content as movies, series, etc.
Write an SQL query to?report the distinct titles of the kid-friendly movies streamed in June 2020.
SELECT DISTINCT title AS title
FROM content AS c
LEFT JOIN tvprogram AS t
ON c.content_id = t.content_id
WHERE MONTH(program_date) = 6 AND YEAR(program_date) = 2020 AND content_type = 'movies' AND kids_content = 'Y'
;
39.?Create a Session Bar Chart
Table:?Sessions
+---------------------+---------+
| Column Name? ? ? ? | Type? ? |
+---------------------+---------+
| session_id? ? ? ? ? | int? ? |
| duration? ? ? ? ? ? | int? ? |
+---------------------+---------+
session_id is the primary key for this table.
duration is the time in seconds that a user has visited the application.
You want to know how long a user visits your application. You decided to create bins of "[0-5>", "[5-10>", "[10-15>" and "15 minutes or more" and count the number of sessions on it.
Write an SQL query to report the (bin, total) in?any?order.
這道題需要注意不能直接用case when，因為case when如果“when”沒有合適的條件的話，不會顯示這一條件的。如果想把分組條件全部顯示出來，則必須要另想辦法。
SELECT?"[0-5>" AS bin, SUM(CASE WHEN duration/60 BETWEEN 0 AND 5 THEN 1 ELSE 0 END) AS total FROM sessions
UNION
SELECT?"[5-10>" AS bin, SUM(CASE WHEN duration/60 BETWEEN 5 AND 10 THEN 1 ELSE 0 END) AS total?FROM sessions
UNION
SELECT?"[10-15>" AS bin, SUM(CASE WHEN duration/60 BETWEEN 10 AND 15 THEN 1 ELSE 0 END) AS total?FROM sessions
UNION
SELECT?"15 or more" AS bin, SUM(CASE WHEN duration/60 >= 15 THEN 1 ELSE 0 END) AS total?FROM sessions
;
40.?Top Travellers
Table:?Users
+---------------+---------+
| Column Name? | Type? ? |
+---------------+---------+
| id? ? ? ? ? ? | int? ? |
| name? ? ? ? ? | varchar |
+---------------+---------+
id is the primary key for this table.
name is the name of the user.
Table:?Rides
+---------------+---------+
| Column Name? | Type? ? |
+---------------+---------+
| id? ? ? ? ? ? | int? ? |
| user_id? ? ? | int? ? |
| distance? ? ? | int? ? |
+---------------+---------+
id is the primary key for this table.
user_id is the id of the user who travelled the distance "distance".
Write an SQL query to?report the distance travelled by each user.
Return the result table ordered by?travelled_distance?in?descending order, if two or more users travelled the same distance, order them by their?name?in?ascending order.
SELECT u.name, IFNULL(travelled_distance, 0) AS?travelled_distance
FROM
(SELECT user_id, SUM(distance) AS?travelled_distance
FROM rides
GROUP BY user_id) AS a
RIGHT JOIN users AS u
ON a.user_id = u.id
ORDER BY?travelled_distance DESC, name
;
這個題要注意的是，join的時候一定要確認好哪個表是base表；還有就是join以后，如果要求顯示每個user的結果，說明即便結果是null，也要顯示出來0（或者說符合實際意義），注意ifnull的使用。
41.?Ads Performance
Table:?Ads
+---------------+---------+
| Column Name? | Type? ? |
+---------------+---------+
| ad_id? ? ? ? | int? ? |
| user_id? ? ? | int? ? |
| action? ? ? ? | enum? ? |
+---------------+---------+
(ad_id, user_id) is the primary key for this table.
Each row of this table contains the ID of an Ad, the ID of a user and the action taken by this user regarding this Ad.
The action column is an ENUM type of ('Clicked', 'Viewed', 'Ignored').
A company is running Ads and wants to calculate the performance of each Ad.
Performance of the Ad is measured using?Click-Through Rate (CTR) where:
Write an SQL query to find the?ctr?of each Ad.
Round?ctrto 2 decimal points.?Order?the result table by?ctr in descending order?and by ad_id in ascending order in case of a tie.
SELECT ad_id, ROUND(IFNULL(SUM(clicked)*100/(SUM(clicked)+SUM(viewed)), 0), 2) AS ctr
FROM
(SELECT ad_id,?
????????????????CASE WHEN action = 'clicked' THEN 1 ELSE 0 END AS clicked,
? ??????????????CASE WHEN action = 'viewed' THEN 1 ELSE 0 END AS viewed
FROM ads) AS a
GROUP BY ad_id
ORDER BY ctr DESC, ad_id
;
42.?Weather Type in Each Country
Table:?Countries
+---------------+---------+
| Column Name? | Type? ? |
+---------------+---------+
| country_id? ? | int? ? |
| country_name? | varchar |
+---------------+---------+
country_id is the primary key for this table.
Each row of this table contains the ID and the name of one country.
Table:?Weather
+---------------+---------+
| Column Name? | Type? ? |
+---------------+---------+
| country_id? ? | int? ? |
| weather_state | varchar |
| day? ? ? ? ? | date? ? |
+---------------+---------+
(country_id, day) is the primary key for this table.
Each row of this table indicates the weather state in a country for one day.
Write an SQL query to find the type of weather in each country for November 2019.
The type of weather is?Cold?if the average?weather_state?is less than or equal 15,?Hot?if the average?weather_state?is greater than or equal 25 and?Warm?otherwise.
SELECT b.country_name, a.weather_type
FROM
(SELECT country_id,?
????????????????CASE WHEN?AVG(weather_state) <= 15 THEN 'Cold'
? ?????????????????????????WHEN?AVG(weather_state) >= 25 THEN 'Hot'
? ??????????????????????????ELSE 'Warm'?
????????????????END AS weather_type
FROM weather
WHERE MONTH(day) = 11 AND YEAR(day) = 2019
GROUP BY country_id) AS a
LEFT JOIN countries AS b
ON a.country_id = b.country_id
;
注意，在應試的情況下，每個column里面輸入的內容最好還是注意大小寫，不然怕機器過不了。
43.?Students and Examinations
Table:?Students
+---------------+---------+
| Column Name? | Type? ? |
+---------------+---------+
| student_id? ? | int? ? |
| student_name? | varchar |
+---------------+---------+
student_id is the primary key for this table.
Each row of this table contains the ID and the name of one student in the school.
Table:?Subjects
+--------------+---------+
| Column Name? | Type? ? |
+--------------+---------+
| subject_name | varchar |
+--------------+---------+
subject_name is the primary key for this table.
Each row of this table contains the name of one subject in the school.
Table:?Examinations
+--------------+---------+
| Column Name? | Type? ? |
+--------------+---------+
| student_id? | int? ? |
| subject_name | varchar |
+--------------+---------+
There is no primary key for this table. It may contain duplicates.
Each student from the Students table takes every course from Subjects table.
Each row of this table indicates that a student with ID student_id attended the exam of subject_name.
Write an SQL query to find the number of times each student attended each exam.
Order the result table by?student_id?and?subject_name.
WITH temp AS (SELECT student_id, student_name, subject_name?FROM students, subjects)
SELECT temp.student_id, temp.student_name, temp.subject_name, IFNULL(a.attended_exams, 0) AS?attended_exams
FROM
(SELECT?student_id, subject_name, COUNT(subject_name) AS?attended_exams
FROM examinations AS e
GROUP BY student_id, subject_name) AS a
RIGHT JOIN temp
ON a.student_id = temp.student_id AND a.subject_name = temp.subject_name
ORDER BY?student_id, subject_name
;
這道題其實有點tricky的地方在于三個表之中，最后一個表是主表，但是存在缺失信息，如果是跟另兩張表挨個join的話，其實中間很有可能會出錯，所以最好是先把兩張信息無缺失的表cross join成一張標準表，然后將最后一張表的信息算出來之后再和標準表join，最后join完select的時候要注意ifnull，最后的order by也不能忘記寫。
