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[introduction to SQL in 10 days] day5+6 consolidated table

2022-06-28 08:25:00 【Ly methane】

175. Combine two tables

 surface : Person

+-------------+---------+
|  Name          |  type      |
+-------------+---------+
| PersonId    | int     |
| FirstName   | varchar |
| LastName    | varchar |
+-------------+---------+
personId  Is the primary key column of the table .
 This table contains some people's  ID  And their last and first names .
+-------------+---------+
|  Name          |  type     |
+-------------+---------+
| AddressId   | int     |
| PersonId    | int     |
| City        | varchar |
| State       | varchar |
+-------------+---------+
addressId  Is the primary key column of the table . Each row of the table contains a  ID = PersonId  Information about people's cities and states .
 Write a SQL Query to report  Person  The last name of each person in the list 、 name 、 Cities and states . If  personId  Your address is not in  Address  In the table , The report is empty   null .

 Input : 
Person surface :
+----------+----------+-----------+
| personId | lastName | firstName |
+----------+----------+-----------+
| 1        | Wang     | Allen     |
| 2        | Alice    | Bob       |
+----------+----------+-----------+
Address surface :
+-----------+----------+---------------+------------+
| addressId | personId | city          | state      |
+-----------+----------+---------------+------------+
| 1         | 2        | New York City | New York   |
| 2         | 3        | Leetcode      | California |
+-----------+----------+---------------+------------+
 Output : 
+-----------+----------+---------------+----------+
| firstName | lastName | city          | state    |
+-----------+----------+---------------+----------+
| Allen     | Wang     | Null          | Null     |
| Bob       | Alice    | New York City | New York |
+-----------+----------+---------------+----------+

Analysis of the answer

Left connection , personId Same connection in one piece

SELECT A.firstName, A.lastName, B.city, B.state 
FROM person A LEFT JOIN address B ON A.personId = B.personId

1581. Customers who enter the store without trading

 surface ：Visits
+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| visit_id    | int     |
| customer_id | int     |
+-------------+---------+
visit_id  Is the primary key of the table . This table contains information about customers who have visited the shopping center .

 surface ：Transactions
+----------------+---------+
| Column Name    | Type    |
+----------------+---------+
| transaction_id | int     |
| visit_id       | int     |
| amount         | int     |
+----------------+---------+
transaction_id  Is the primary key of this table . This table contains  visit_id  Information on transactions conducted during .

 Some customers may visit the shopping center but do not trade . Please write a  SQL  Inquire about , To find these customers  ID , And the number of times they only patronize and don't trade .
 Return to   In any order   Sorted result table .
+-------------+----------------+
| customer_id | count_no_trans |
+-------------+----------------+
| 54          | 2              |
| 30          | 1              |
| 96          | 1              |
+-------------+----------------+

Analysis of the answer

First, find out through the joint statement that there has been no transaction visit,
Then, these query results are processed according to customer_id grouping , Statistics customer_id And the number of each group

SELECT customer_id, count(visit_id) count_no_trans
FROM (
    SELECT distinct A.visit_id, A.customer_id 
    FROM visits A LEFT JOIN transactions B ON A.visit_id = B.visit_id
    WHERE B.amount IS NULL) C
GROUP BY C.customer_id

You can also check directly visits Not in transactions The data in , Then calculate in groups customer_id The same number

SELECT customer_id, count(visit_id) count_no_trans 
FROM visits
WHERE visit_id NOT IN (SELECT distinct visit_id FROM transactions)
GROUP BY customer_id```

1148. Article browsing I

Views  surface ：
+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| article_id    | int     |
| author_id     | int     |
| viewer_id     | int     |
| view_date     | date    |
+---------------+---------+
 This table has no primary key , Therefore, there may be duplicate lines .
 Each row of this table indicates that someone browsed an article by an author on a certain day .
 Please note that , Of the same person  author_id  and  viewer_id  It's the same .
 Please write one  SQL  Query to find all authors who have browsed their own articles , Results according to  id  Ascending order .
Views  surface ：
+------------+-----------+-----------+------------+
| article_id | author_id | viewer_id | view_date  |
+------------+-----------+-----------+------------+
| 1          | 3         | 5         | 2019-08-01 |
| 1          | 3         | 6         | 2019-08-02 |
| 2          | 7         | 7         | 2019-08-01 |
| 2          | 7         | 6         | 2019-08-02 |
| 4          | 7         | 1         | 2019-07-22 |
| 3          | 4         | 4         | 2019-07-21 |
| 3          | 4         | 4         | 2019-07-21 |
+------------+-----------+-----------+------------+
 Result sheet ：
+------+
| id   |
+------+
| 4    |
| 7    |
+------+

Analysis of the answer

Inquire about author_id and viewer_id equal

SELECT distinct author_id id 
FROM views
WHERE author_id = viewer_id
ORDER BY id

197. The rising temperature

 surface ： Weather
+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| recordDate    | date    |
| temperature   | int     |
+---------------+---------+
id  It's the primary key of this table , The table contains temperature information for a specific date 
 Write a  SQL  Inquire about , To find out （ Yesterday's ） Of all dates with a higher temperature  id .
 Return results   Order is not required  .

 Input ：
Weather  surface ：
+----+------------+-------------+
| 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  |
+----+
 explain ：
2015-01-02  The temperature is higher than the day before （10 -> 25）
2015-01-04  The temperature is higher than the day before （20 -> 30）

Analysis of the answer

DATEDIFF stay MySQL In the middle of the day

DATEDIFF(date1,date2)   #  return date1 - date2  Days of difference

Self join , The day after the query and the temperature is high

SELECT A.id 
FROM weather A, Weather B 
WHERE A.Temperature > B.Temperature  AND DATEDIFF(A.recordDate, B.recordDate) = 1

607. Salesperson

 surface : SalesPerson
+-----------------+---------+
| Column Name     | Type    |
+-----------------+---------+
| sales_id        | int     |
| name            | varchar |
| salary          | int     |
| commission_rate | int     |
| hire_date       | date    |
+-----------------+---------+
sales_id  Is the primary key column of the table . Each row of the table shows the salesperson's name and  ID , And their wages 、 Commission rate and date of employment .
 
 surface : Company
+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| com_id      | int     |
| name        | varchar |
| city        | varchar |
+-------------+---------+
com_id  Is the primary key column of the table . Each row of the table indicates the name of the company and  ID , And the city where the company is located .
 
 surface : 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 of the table .
com_id  yes  Company  In the table  com_id  The foreign key .
sales_id  It's from the salesperson table  sales_id  The foreign key .
 Each row of the table contains the information of an order . This includes the company's  ID 、 Salesperson's  ID 、 Date of order and amount paid .
 
 Write a SQL Inquire about , The report has nothing to do with the name  “RED”  The names of all salespeople for orders related to the company .

 Input ：
SalesPerson  surface :
+----------+------+--------+-----------------+------------+
| 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  surface :
+--------+--------+----------+
| com_id | name   | city     |
+--------+--------+----------+
| 1      | RED    | Boston   |
| 2      | ORANGE | New York |
| 3      | YELLOW | Boston   |
| 4      | GREEN  | Austin   |
+--------+--------+----------+
Orders  surface :
+----------+------------+--------+----------+--------+
| 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 |
+------+
 According to the table  orders  Order in  '3'  and  '4' , It's easy to see that there is only  'John'  and  'Pam'  Two salesmen once told the company  'RED'  Sold .
 So we need output meters  salesperson  The names of all the others in .

Analysis of the answer

First in Company Find the company in the table RED The company ID
And then Orders Find out in the table that the company's order involves sales_id
Last query SalesPerson In the table Query the names of people who are not in the involved list

SELECT A.name 
FROM salesperson A
WHERE A.sales_id NOT IN(SELECT C.sales_id as id
						FROM orders C
						WHERE C.com_id IN(SELECT B.com_id 
										  FROM Company B
										  WHERE B.name = 'RED'))