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이것저것 참고용

쿼리 테스트

도토리즈 2023. 6. 15. 23:23

- create , insert 쿼리 

CREATE TABLE Customers (
    customer_id INT PRIMARY KEY,
    customer_name VARCHAR(50),
    email VARCHAR(100),
    phone_number VARCHAR(15)
);

INSERT INTO Customers (customer_id, customer_name, email, phone_number)
VALUES
    (1, 'John Doe', 'john.doe@example.com', '123-456-7890'),
    (2, 'Jane Smith', 'jane.smith@example.com', '987-654-3210'),
    (3, 'Bob Johnson', 'bob.johnson@example.com', '555-555-5555');

CREATE TABLE Orders (
    order_id INT PRIMARY KEY,
    customer_id INT,
    order_date DATE,
    total_amount DECIMAL(10, 2),
    product_id INT,
    FOREIGN KEY (customer_id) REFERENCES Customers(customer_id),
    FOREIGN KEY (product_id) REFERENCES Products(product_id)
);

INSERT INTO Orders (order_id, customer_id, order_date, total_amount, product_id)
VALUES
    (1, 1, '2023-06-01', 100.00, 1),
    (2, 2, '2023-06-02', 50.00, 2),
    (3, 1, '2023-06-03', 75.00, 1),
    (4, 3, '2023-06-04', 200.00, 3);

CREATE TABLE Products (
    product_id INT PRIMARY KEY,
    product_name VARCHAR(50),
    price DECIMAL(10, 2)
);

ALTER TABLE Products
ADD category_id INT;

INSERT INTO Products (product_id, product_name, price)
VALUES
    (1, 'Product A', 10.00),
    (2, 'Product B', 20.00),
    (3, 'Product C', 30.00);
- 테스트쿼리 1
SELECT * 
    FROM (
        SELECT Customers.customer_id, Customers.customer_name, Customers.email, 
        Customers.phone_number, Orders.order_id, Orders.order_date, Orders.total_amount, Orders.product_id
        FROM Customers
         JOIN Orders ON Customers.customer_id = Orders.customer_id
    ) AS CustomerOrders
     JOIN Products ON CustomerOrders.product_id = Products.product_id;
     
     
 - 테스트쿼리2   
 SELECT *
    FROM (
        SELECT *
        FROM Customers
        INNER JOIN Orders ON Customers.customer_id = Orders.customer_id
    ) AS CustomerOrders
    INNER JOIN Products ON  CustomerOrders.product_id = Products.product_id ;

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