[introduction to SQL for 10 days] day4 Combined Query & specified selection

1965. Employees who lost information

 surface : Employees
+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| employee_id | int     |
| name        | varchar |
+-------------+---------+
employee_id  It's the primary key of this table . Each line represents the employee's id  And his name .
 surface : Salaries
+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| employee_id | int     |
| salary      | int     |
+-------------+---------+
employee_id is  The primary key of this table . Each line represents the employee's id  And his salary .

 Write a query statement , Find all   Employees   full name   lost , Or employee's   Salary information   Lost employees id.
 Return the of these employees id  employee_id ,  Sort from small to large  .

Analysis of the answer

League table query , Full connection , however mysql It's not worth money , All can be left connected and connected together
Union： Combine two result sets , Do not include repeating lines , At the same time, sort the default rules ;

SELECT A.employee_id FROM employees A LEFT JOIN salaries B ON A.employee_id = B.employee_id
WHERE B.salary IS NULL
UNION
SELECT B.employee_id FROM employees A RIGHT JOIN salaries B ON A.employee_id = B.employee_id
WHERE A.name IS NULL
ORDER BY employee_id

1795. The price of each product in different stores

 surface ：Products
+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| product_id  | int     |
| store1      | int     |
| store2      | int     |
| store3      | int     |
+-------------+---------+
 The primary key of this table is product_id（ product Id）.
 Each line stores this product in different stores store1, store2, store3 The price of . If this product is not sold in the store , The value will be null.
 Please refactor  Products  surface , Check the price of each product in different stores ,
 Make the output format change to (product_id, store, price) . If this product is not sold in the store , This line is not output .
 In the output result table   The order is not required  .
 Input ：
Products table:
+------------+--------+--------+--------+
| product_id | store1 | store2 | store3 |
+------------+--------+--------+--------+
| 0          | 95     | 100    | 105    |
| 1          | 70     | null   | 80     |
+------------+--------+--------+--------+
 Output ：
+------------+--------+-------+
| product_id | store  | price |
+------------+--------+-------+
| 0          | store1 | 95    |
| 0          | store2 | 100   |
| 0          | store3 | 105   |
| 1          | store1 | 70    |
| 1          | store3 | 80    |
+------------+--------+-------+

Analysis of the answer

For row to column groupby, Column to row union all
union all and union The difference is that union all No weight removal

sumif

SELECT product_id, 'store1' store,store1 price 
FROM products
WHERE store1 IS NOT NULL
UNION ALL
SELECT product_id, 'store2' store,store2 price 
FROM products
WHERE store2 IS NOT NULL
UNION ALL
SELECT product_id, 'store3' store,store3 store 
FROM products
WHERE store3 IS NOT NULL

608. Tree node

 Given a table  tree,id  Is the number of the tree node , p_id  Is its parent node  id .
+----+------+
| id | p_id |
+----+------+
| 1  | null |
| 2  | 1    |
| 3  | 1    |
| 4  | 2    |
| 5  | 2    |
+----+------+
 Each node in the tree belongs to one of the following three types ：
- leaf ： If this node does not have any child nodes .
-- root ： If this node is the root of the whole tree , That is, there is no parent node .
-- Internal node ： If this node is neither a leaf node nor a root node .
 Write a query statement , Output the number and type of all nodes , And sort the results according to the node number . The result of the above example is ：
+----+------+
| id | Type |
+----+------+
| 1  | Root |
| 2  | Inner|
| 3  | Leaf |
| 4  | Leaf |
| 5  | Leaf |
+----+------+

Analysis of the answer

not in And null value , If not in There are null value , Then go straight back to null The corresponding result is false
ordinary case Several usages of the sentence

CASE WHEN condition THEN result ELSE result END

CASE WHEN p_id IS NULL THEN 'Root'
	 WHEN id NOT IN(SELECT p_id FROM tree WHERE p_id IS NOT NULL) THEN 'Leaf' ELSE 'Inner' END
	 
CASE SCORE WHEN 'A' THEN ' optimal ' ELSE ' fail, ' END

answer ：

SELECT 
	id, 
	CASE WHEN p_id IS NULL THEN 'Root'
	 	 WHEN id NOT IN(SELECT p_id FROM tree WHERE p_id IS NOT NULL) THEN 'Leaf' ELSE 'Inner' END AS 'type'
FROM tree

176. The second highest salary

Employee  surface ：
+-------------+------+
| Column Name | Type |
+-------------+------+
| id          | int  |
| salary      | int  |
+-------------+------+
id  It's the primary key of this table .
 Each row of the table contains the salary information of the employee .
 Write a  SQL  Inquire about , Get and return  Employee  The second highest salary on the list  . If there is no second highest salary , The query should return  null .
 Input ：
Employee  surface ：
+----+--------+
| id | salary |
+----+--------+
| 1  | 100    |
| 2  | 200    |
| 3  | 300    |
+----+--------+
 Output ：
+---------------------+
| SecondHighestSalary |
+---------------------+
| 200                 |
+---------------------+

Analysis of the answer

limit usage ：

select * from tableName limit i,n
# tableName： Table name 
# i： Is the index value of the query result ( The default from the 0 Start ), When i=0 It can be omitted i
# n： Number returned for query results 
# i And n Use English commas between "," separate 

keyword distinct duplicate removal

SELECT DISTINCT name from A

Query the second largest , First query all , Then go heavy. , Then sort in descending order , And get limit Get the second
If the above operation does not find a value , Then return null, Then name the query content

SELECT IFNULL(
    (SELECT DISTINCT salary
	FROM employee 
	ORDER BY salary DESC
	LIMIT 1, 1), null) SecondHighestSalary 

summary ：

1. MySQL There is no external connection , Connect left and right + union all
2. Break up the whole line into several lines union, Merge multiple rows into a whole row group by
3. If not in There are null value , Then go straight back to null The corresponding result is false, It doesn't really judge whether there is in the collection
4. case Sentence usage CASE WHEN condition THEN result1 ELSE result2 END
5. DISTINCT duplicate removal SELECT DISTINCT name from A
6. LIMIT usage take i - n, from 0 Start ： SELECT * FROM table limit i,n