[SQL brush questions] DAY14----Special exercise on using subqueries in SQL

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1. SQL uses subqueries

(1) Subquery

A subquery is a query nested in another statement, such as: select , insert, update, delete

(2) Nested subqueries

Subqueries can be nested within another subquery, SQL Server supports up to 32 nesting levels

(3) Related subroutines

① A correlated subquery is a subquery that uses the values ​​of the outer query . i.e. it depends on the value of the outer query

② Correlated subqueries cannot be executed independently as simple subqueries

③ Repeatedly execute the correlated subquery once for each row evaluated by the outer query. Correlated subqueries are also known as repeating subqueries .

2. Brush questions

1. Brush Question 1

(1) Topics

Question: Return a list of customers who purchased products for $10 or more

Description: OrderItems represents the order item table, containing the fields order number: order_num, order price: item_price; the Orders table represents the order information table, including customer id: cust_id and order number: order_num

(2) Example

输入：

DROP TABLE IF EXISTS `OrderItems`;

  CREATE TABLE IF NOT EXISTS `OrderItems`(

    order_num VARCHAR(255) NOT NULL COMMENT '商品订单号',

    item_price INT(16) NOT NULL COMMENT '售出价格'

  );

  INSERT `OrderItems` VALUES ('a1',10),('a2',1),('a2',1),('a4',2),('a5',5),('a2',1),('a7',7);


  DROP TABLE IF EXISTS `Orders`;

  CREATE TABLE IF NOT EXISTS `Orders`(

    order_num VARCHAR(255) NOT NULL COMMENT '商品订单号',

    cust_id VARCHAR(255) NOT NULL COMMENT '顾客id'

  );

  INSERT `Orders` VALUES ('a1','cust10'),('a2','cust1'),('a2','cust1'),('a4','cust2'),('a5','cust5'),('a2','cust1'),('a7','cust7');


输出：


cust10

(3) Code

select cust_idfrom Orderswhere order_num in (

    select order_num

    from OrderItems

    where item_price >=10

)

(4) Operation result

​

2. Brush question 2

(1) Topics

Topic: Determine which orders purchased the product with prod_id BR01 (1)

Description: The table OrderItems represents the order item information table, prod_id is the product id; the Orders table represents the order table, and the cust_id represents the customer id and the order date order_date

(2) Example

输入：

DROP TABLE IF EXISTS `OrderItems`;

  CREATE TABLE IF NOT EXISTS `OrderItems`(

    prod_id VARCHAR(255) NOT NULL COMMENT '产品id',

    order_num VARCHAR(255) NOT NULL COMMENT '商品订单号'

  );

  INSERT `OrderItems` VALUES ('BR01','a0001'),('BR01','a0002'),('BR02','a0003'),('BR02','a0013');


  DROP TABLE IF EXISTS `Orders`;

  CREATE TABLE IF NOT EXISTS `Orders`(

    order_num VARCHAR(255) NOT NULL COMMENT '商品订单号',

    cust_id VARCHAR(255) NOT NULL COMMENT '顾客id',

    order_date TIMESTAMP NOT NULL COMMENT '下单时间'

  );

  INSERT `Orders` VALUES ('a0001','cust10','2022-01-01 00:00:00'),('a0002','cust1','2022-01-01 00:01:00'),('a0003','cust1','2022-01-02 00:00:00'),('a0013','cust2','2022-01-01 00:20:00');


输出：


cust10|2022-01-01 00:00:00

cust1|2022-01-01 00:01:00

(3) Code

select cust_id,order_datefrom Orderswhere order_num in

(select order_numfrom OrderItemswhere prod_id = 'BR01')

(4) Operation result

​

3. Brush question three

(1) Topics

Title: Return the emails of all customers who purchased products with prod_id BR01 (1)

Description: You want to know the date of ordering BR01 products, there is a table OrderItems representing order item information table, prod_id is product id; Orders table represents order table, cust_id represents customer id and order date order_date; Customers table contains cust_email customer email and cust_id customer id

 

(2) Example

输入：

DROP TABLE IF EXISTS `OrderItems`;

  CREATE TABLE IF NOT EXISTS `OrderItems`(

    prod_id VARCHAR(255) NOT NULL COMMENT '产品id',

    order_num VARCHAR(255) NOT NULL COMMENT '商品订单号'

  );

  INSERT `OrderItems` VALUES ('BR01','a0001'),('BR01','a0002'),('BR02','a0003'),('BR02','a0013');


  DROP TABLE IF EXISTS `Orders`;

  CREATE TABLE IF NOT EXISTS `Orders`(

    order_num VARCHAR(255) NOT NULL COMMENT '商品订单号',

    cust_id VARCHAR(255) NOT NULL COMMENT '顾客id',

    order_date TIMESTAMP NOT NULL COMMENT '下单时间'

  );

  INSERT `Orders` VALUES ('a0001','cust10','2022-01-01 00:00:00'),('a0002','cust1','2022-01-01 00:01:00'),('a0003','cust1','2022-01-02 00:00:00'),('a0013','cust2','2022-01-01 00:20:00');

DROP TABLE IF EXISTS `Customers`;CREATE TABLE IF NOT EXISTS `Customers`(

    cust_id VARCHAR(255) NOT NULL COMMENT '顾客id',

    cust_email VARCHAR(255) NOT NULL COMMENT '顾客email'

  );INSERT `Customers` VALUES ('cust10','[email protected]'),('cust1','[email protected]'),('cust2','[email protected]');


输出：


[email protected]

[email protected]

(3) Code

select

    Customers.cust_emailfrom

    Ordersleft join OrderItems on OrderItems.order_num = Orders.order_numleft join Customers on Customers.cust_id = Orders.cust_idwhere

    OrderItems.prod_id = "BR01"

(4) Operation result

4. Brush Question 4

(1) Topics

Question: Return the total amount of each customer's different orders

Description: We need a list of customer IDs containing the total amount they have ordered. The OrderItems table represents order information, and the OrderItems table has the order number: order_num, the selling price of the item: item_price, and the quantity of the item: quantity.

 

(2) Example

输入：

DROP TABLE IF EXISTS `OrderItems`;CREATE TABLE IF NOT EXISTS `OrderItems`(

order_num VARCHAR(255) NOT NULL COMMENT '商品订单号',

item_price INT(16) NOT NULL COMMENT '售出价格',

quantity INT(16) NOT NULL COMMENT '商品数量'

);INSERT `OrderItems` VALUES ('a0001',10,105),('a0002',1,1100),('a0002',1,200),('a0013',2,1121),('a0003',5,10),('a0003',1,19),('a0003',7,5);

DROP TABLE IF EXISTS `Orders`;CREATE TABLE IF NOT EXISTS `Orders`(

  order_num VARCHAR(255) NOT NULL COMMENT '商品订单号',

  cust_id VARCHAR(255) NOT NULL COMMENT '顾客id'

);INSERT `Orders` VALUES ('a0001','cust10'),('a0003','cust1'),('a0013','cust2');


输出：


cust2|2242.000

cust10|1050.000

cust1|104.000

(3) Code

SELECT 

    o.cust_id,

    sum(oi.item_price * oi.quantity) AS total_orderedFROM Orders o JOIN OrderItems oi

    USING (order_num)GROUP BY o.cust_idORDER BY total_ordered DESC

(4) Operation result

 

5. Brush Question 5

(1) Topics

Question: Retrieve all product names and corresponding sales totals from the Products table

Description: Retrieve all product names in the Products table: prod_name, product id: prod_id

OrderItems represents the order item table, order product: prod_id, sold quantity: quantity

​

 

(2) Example

enter:

DROP TABLE IF EXISTS `Products`;CREATE TABLE IF NOT EXISTS `Products` (

`prod_id` VARCHAR(255) NOT NULL COMMENT '产品 ID',

`prod_name` VARCHAR(255) NOT NULL COMMENT '产品名称'

);INSERT INTO `Products` VALUES ('a0001','egg'),

('a0002','sockets'),

('a0013','coffee'),

('a0003','cola');

DROP TABLE IF EXISTS `OrderItems`;CREATE TABLE IF NOT EXISTS `OrderItems`(

prod_id VARCHAR(255) NOT NULL COMMENT '产品id',

quantity INT(16) NOT NULL COMMENT '商品数量'

);INSERT `OrderItems` VALUES ('a0001',105),('a0002',1100),('a0002',200),('a0013',1121),('a0003',10),('a0003',19),('a0003',5);


输出：


egg|105.000

sockets|1300.000

coffee|1121.000

cola|34.000

(3) Code

select

  prod_name,

  sum(quantity) as quant_soldfrom

  Products,

  OrderItemswhere

  OrderItems.prod_id = Products.prod_idgroup by

  prod_name

(4) Operation result

​

This article is the fourteenth day of SQL brushing questions

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