今天也是菜味十足，5題中有3題都要去爬文了解才能完成，不過晚上終於可以狂歡看球了~

197. Rising Temperature

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| recordDate    | date    |
| temperature   | int     |
+---------------+---------+
id is the primary key for this table.
This table contains information about the temperature on a certain day.

Write an SQL query to find all dates' Id with higher temperatures compared to its previous dates (yesterday).

Return the result table in any order.

The query result format is in the following example.

Example 1:

Input: 
Weather table:
+----+------------+-------------+
| id | recordDate | temperature |
+----+------------+-------------+
| 1  | 2015-01-01 | 10          |
| 2  | 2015-01-02 | 25          |
| 3  | 2015-01-03 | 20          |
| 4  | 2015-01-04 | 30          |
+----+------------+-------------+
Output: 
+----+
| id |
+----+
| 2  |
| 4  |
+----+
Explanation: 
In 2015-01-02, the temperature was higher than the previous day (10 -> 25).
In 2015-01-04, the temperature was higher than the previous day (20 -> 30).

這一題要找出比前一天溫度還高的id，
比如id2:01-02是25度比id1:01-01的10度高，所以要輸出id'2'
，這邊學會了DATEDIFF的用法，-1的意思就是昨天，
可以到w3s去試試變化。

SELECT  W2.id
FROM Weather W1 INNER JOIN Weather W2 #只用join也會過
ON DATEDIFF(W1.recordDate, W2.recordDate)=-1
AND W2.temperature > W1.temperature;

607. Sales Person

+-----------------+---------+
| Column Name     | Type    |
+-----------------+---------+
| sales_id        | int     |
| name            | varchar |
| salary          | int     |
| commission_rate | int     |
| hire_date       | date    |
+-----------------+---------+
sales_id is the primary key column for this table.
Each row of this table indicates the name and the ID of a salesperson alongside their salary, commission rate, and hire date.
 
Table: Company

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| com_id      | int     |
| name        | varchar |
| city        | varchar |
+-------------+---------+
com_id is the primary key column for this table.
Each row of this table indicates the name and the ID of a company and the city in which the company is located.

Table: Orders

+-------------+------+
| Column Name | Type |
+-------------+------+
| order_id    | int  |
| order_date  | date |
| com_id      | int  |
| sales_id    | int  |
| amount      | int  |
+-------------+------+
order_id is the primary key column for this table.
com_id is a foreign key to com_id from the Company table.
sales_id is a foreign key to sales_id from the SalesPerson table.
Each row of this table contains information about one order. This includes the ID of the company, the ID of the salesperson, the date of the order, and the amount paid.

Write an SQL query to report the names of all the salespersons who did not have any orders related to the company with the name "RED".

Return the result table in any order.

The query result format is in the following example.

Example 1:

Input: 
SalesPerson table:
+----------+------+--------+-----------------+------------+
| sales_id | name | salary | commission_rate | hire_date  |
+----------+------+--------+-----------------+------------+
| 1        | John | 100000 | 6               | 4/1/2006   |
| 2        | Amy  | 12000  | 5               | 5/1/2010   |
| 3        | Mark | 65000  | 12              | 12/25/2008 |
| 4        | Pam  | 25000  | 25              | 1/1/2005   |
| 5        | Alex | 5000   | 10              | 2/3/2007   |
+----------+------+--------+-----------------+------------+
Company table:
+--------+--------+----------+
| com_id | name   | city     |
+--------+--------+----------+
| 1      | RED    | Boston   |
| 2      | ORANGE | New York |
| 3      | YELLOW | Boston   |
| 4      | GREEN  | Austin   |
+--------+--------+----------+
Orders table:
+----------+------------+--------+----------+--------+
| order_id | order_date | com_id | sales_id | amount |
+----------+------------+--------+----------+--------+
| 1        | 1/1/2014   | 3      | 4        | 10000  |
| 2        | 2/1/2014   | 4      | 5        | 5000   |
| 3        | 3/1/2014   | 1      | 1        | 50000  |
| 4        | 4/1/2014   | 1      | 4        | 25000  |
+----------+------------+--------+----------+--------+
Output: 
+------+
| name |
+------+
| Amy  |
| Mark |
| Alex |
+------+
Explanation: 
According to orders 3 and 4 in the Orders table, it is easy to tell that only salesperson John and Pam have sales to company RED, so we report all the other names in the table salesperson.

這一題要列出不在RED公司工作過的員工姓名，我的思路如下：
確認員工姓名在SalesPerson表中，再開始找其他表的關聯性。
1.首先找到RED公司，公司在Company表的name中，com_id為1
2.com_id為1的值出現在Orders表中(有兩筆)，而sales_id能關聯到SalesPerson表
3.sales_id(1、4)兩位有在RED公司工作過
4.排除掉兩位員工後，剩下的人就是答案

SELECT S.name
FROM SalesPerson S
WHERE S.Name NOT IN
    (SELECT S.name 
    FROM Orders O
    INNER JOIN Company C
    ON C.com_id = O.com_id 
    LEFT JOIN SalesPerson S ON S.sales_id =O.sales_id
    WHERE C.Name LIKE "RED")

1141. User Activity for the Past 30 Days I

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| user_id       | int     |
| session_id    | int     |
| activity_date | date    |
| activity_type | enum    |
+---------------+---------+
There is no primary key for this table, it may have duplicate rows.
The activity_type column is an ENUM of type ('open_session', 'end_session', 'scroll_down', 'send_message').
The table shows the user activities for a social media website. 
Note that each session belongs to exactly one user.
 
Write an SQL query to find the daily active user count for a period of 30 days ending 2019-07-27 inclusively. A user was active on someday if they made at least one activity on that day.

Return the result table in any order.

The query result format is in the following example.

Example 1:

Input: 
Activity table:
+---------+------------+---------------+---------------+
| user_id | session_id | activity_date | activity_type |
+---------+------------+---------------+---------------+
| 1       | 1          | 2019-07-20    | open_session  |
| 1       | 1          | 2019-07-20    | scroll_down   |
| 1       | 1          | 2019-07-20    | end_session   |
| 2       | 4          | 2019-07-20    | open_session  |
| 2       | 4          | 2019-07-21    | send_message  |
| 2       | 4          | 2019-07-21    | end_session   |
| 3       | 2          | 2019-07-21    | open_session  |
| 3       | 2          | 2019-07-21    | send_message  |
| 3       | 2          | 2019-07-21    | end_session   |
| 4       | 3          | 2019-06-25    | open_session  |
| 4       | 3          | 2019-06-25    | end_session   |
+---------+------------+---------------+---------------+
Output: 
+------------+--------------+ 
| day        | active_users |
+------------+--------------+ 
| 2019-07-20 | 2            |
| 2019-07-21 | 2            |
+------------+--------------+ 
Explanation: Note that we do not care about days with zero active users.

這一題希望我們列出一個月中的用戶活躍數，06-25的用戶就會被排除掉，
07-20有兩個用戶活動
07-21也有兩個用戶活動
要注意ID:2看起來是跨日活動，會分開算。

SELECT activity_date AS day , COUNT(DISTINCT(user_id)) AS active_users
FROM Activity
WHERE activity_date > "2019-06-27" AND 
        activity_date <= "2019-07-27"
GROUP BY activity_date
ORDER BY activity_date ASC

1693. Daily Leads and Partners

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| date_id     | date    |
| make_name   | varchar |
| lead_id     | int     |
| partner_id  | int     |
+-------------+---------+
This table does not have a primary key.
This table contains the date and the name of the product sold and the IDs of the lead and partner it was sold to.
The name consists of only lowercase English letters.

Write an SQL query that will, for each date_id and make_name, return the number of distinct lead_id's and distinct partner_id's.

Return the result table in any order.

The query result format is in the following example.

Example 1:

Input: 
DailySales table:
+-----------+-----------+---------+------------+
| date_id   | make_name | lead_id | partner_id |
+-----------+-----------+---------+------------+
| 2020-12-8 | toyota    | 0       | 1          |
| 2020-12-8 | toyota    | 1       | 0          |
| 2020-12-8 | toyota    | 1       | 2          |
| 2020-12-7 | toyota    | 0       | 2          |
| 2020-12-7 | toyota    | 0       | 1          |
| 2020-12-8 | honda     | 1       | 2          |
| 2020-12-8 | honda     | 2       | 1          |
| 2020-12-7 | honda     | 0       | 1          |
| 2020-12-7 | honda     | 1       | 2          |
| 2020-12-7 | honda     | 2       | 1          |
+-----------+-----------+---------+------------+
Output: 
+-----------+-----------+--------------+-----------------+
| date_id   | make_name | unique_leads | unique_partners |
+-----------+-----------+--------------+-----------------+
| 2020-12-8 | toyota    | 2            | 3               |
| 2020-12-7 | toyota    | 1            | 2               |
| 2020-12-8 | honda     | 2            | 2               |
| 2020-12-7 | honda     | 3            | 2               |
+-----------+-----------+--------------+-----------------+
Explanation: 
For 2020-12-8, toyota gets leads = [0, 1] and partners = [0, 1, 2] while honda gets leads = [1, 2] and partners = [1, 2].
For 2020-12-7, toyota gets leads = [0] and partners = [1, 2] while honda gets leads = [0, 1, 2] and partners = [1, 2].

這一題是列出客戶總數，比如toyota 12-08的客戶有lead_id(0、1)、partner_id(0、1、2)
結果就會是unique_leads 2位，unique_partners 3位，依此類推，
因為有用COUNT函式，所以要用GROUP BY排序，
我去GOOGLE找看有沒有跟題目要求的Output完全一樣那種寫法，但沒找到，總之先求PASS了。

SELECT date_id, make_name, 
    COUNT(distinct lead_id) unique_leads,
    COUNT(distinct partner_id) unique_partners
FROM DailySales
GROUP BY date_id,make_name
#OUTPUT會是下面這樣，但也能過關。
/*
| date_id    | make_name | unique_leads | unique_partners |
| ---------- | --------- | ------------ | --------------- |
| 2020-12-07 | honda     | 3            | 2               |
| 2020-12-07 | toyota    | 1            | 2               |
| 2020-12-08 | honda     | 2            | 2               |
| 2020-12-08 | toyota    | 2            | 3               |
*/

1729. Find Followers Count

+-------------+------+
| Column Name | Type |
+-------------+------+
| user_id     | int  |
| follower_id | int  |
+-------------+------+
(user_id, follower_id) is the primary key for this table.
This table contains the IDs of a user and a follower in a social media app where the follower follows the user.
 
Write an SQL query that will, for each user, return the number of followers.

Return the result table ordered by user_id in ascending order.

The query result format is in the following example.

Example 1:

Input: 
Followers table:
+---------+-------------+
| user_id | follower_id |
+---------+-------------+
| 0       | 1           |
| 1       | 0           |
| 2       | 0           |
| 2       | 1           |
+---------+-------------+
Output: 
+---------+----------------+
| user_id | followers_count|
+---------+----------------+
| 0       | 1              |
| 1       | 1              |
| 2       | 2              |
+---------+----------------+
Explanation: 
The followers of 0 are {1}
The followers of 1 are {0}
The followers of 2 are {0,1}

統計追蹤者人數，COUNT(DISTINCT(follower_id))的意思是總計不重複的值，
記得要用GROUP BY排序

SELECT user_id , COUNT(DISTINCT(follower_id)) AS followers_count
FROM Followers
GROUP BY user_id