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코딩은 GPT가 시키는것 #8

Closed choi4624 closed 1 year ago

choi4624 commented 1 year ago

이 영상과 같이 시청하세요 https://www.youtube.com/watch?v=5yquJa2x3Ko

User spring DTO to mapping example ChatGPT Sure, here's an example of mapping a Spring Data Transfer Object (DTO) to a domain model:

Let's assume we have a simple DTO class called UserDTO:

arduino Copy code public class UserDTO { private String firstName; private String lastName; private String email; // getters and setters } And we have a corresponding domain model class called User:

arduino Copy code public class User { private Long id; private String firstName; private String lastName; private String email; // getters and setters } To map UserDTO to User, we can use a mapping library such as MapStruct. First, we need to create a mapper interface that defines how the mapping should be performed:

scss Copy code @Mapper(componentModel = "spring") public interface UserMapper {

UserDTO toDto(User user);

User toEntity(UserDTO userDto);

} The toDto method maps a User object to a UserDTO object, while the toEntity method maps a UserDTO object to a User object.

To use the mapper, we can inject it into a Spring service or controller class and call its methods:

kotlin Copy code @Service public class UserService {

private final UserRepository userRepository;
private final UserMapper userMapper;

public UserService(UserRepository userRepository, UserMapper userMapper) {
    this.userRepository = userRepository;
    this.userMapper = userMapper;
}

public UserDTO createUser(UserDTO userDto) {
    User user = userMapper.toEntity(userDto);
    user = userRepository.save(user);
    return userMapper.toDto(user);
}

} In this example, the createUser method first maps the UserDTO object to a User object using the toEntity method, saves the User object to the database using a UserRepository, and then maps the saved User object back to a UserDTO object using the toDto method before returning it.

That's a basic example of how to map a Spring DTO to a domain model using MapStruct. Of course, there are other mapping libraries and approaches you can use depending on your requirements and preferences.

User is there other example 'without' using Mapstruct? ChatGPT Yes, here's an example of mapping a Spring Data Transfer Object (DTO) to a domain model without using MapStruct:

Let's assume we have a simple DTO class called UserDTO:

arduino Copy code public class UserDTO { private String firstName; private String lastName; private String email; // getters and setters } And we have a corresponding domain model class called User:

arduino Copy code public class User { private Long id; private String firstName; private String lastName; private String email; // getters and setters } To map UserDTO to User, we can create a utility class with static methods that perform the mapping. For example:

scss Copy code public class UserMapper {

public static User toUser(UserDTO userDto) {
    User user = new User();
    user.setFirstName(userDto.getFirstName());
    user.setLastName(userDto.getLastName());
    user.setEmail(userDto.getEmail());
    return user;
}

public static UserDTO toUserDto(User user) {
    UserDTO userDto = new UserDTO();
    userDto.setFirstName(user.getFirstName());
    userDto.setLastName(user.getLastName());
    userDto.setEmail(user.getEmail());
    return userDto;
}

} The toUser method maps a UserDTO object to a User object, while the toUserDto method maps a User object to a UserDTO object.

To use the UserMapper class, we can simply call its static methods from a Spring service or controller class:

java Copy code @Service public class UserService {

private final UserRepository userRepository;

public UserService(UserRepository userRepository) {
    this.userRepository = userRepository;
}

public UserDTO createUser(UserDTO userDto) {
    User user = UserMapper.toUser(userDto);
    user = userRepository.save(user);
    return UserMapper.toUserDto(user);
}

} In this example, the createUser method first maps the UserDTO object to a User object using the toUser method, saves the User object to the database using a UserRepository, and then maps the saved User object back to a UserDTO object using the toUserDto method before returning it.

That's a basic example of how to map a Spring DTO to a domain model without using a mapping library like MapStruct. Note that this approach can become tedious and error-prone if you have many fields or complex mappings, so using a mapping library is generally recommended for more complex scenarios.

User can public class UserDTO conver as records? ChatGPT Yes, you can convert the UserDTO class to a record in Java 16 or later. A record is a new feature introduced in Java 16 that provides a concise way to declare a class with immutable state.

To convert the UserDTO class to a record, you can simply replace the class keyword with the record keyword, like this:

arduino Copy code public record UserDTO(String firstName, String lastName, String email) {} This declares a UserDTO record with three components: firstName, lastName, and email. The record keyword automatically generates a constructor and getters for the components, as well as implementations of the equals, hashCode, and toString methods.

You can use the UserDTO record in the same way as the original class, for example:

java Copy code UserDTO userDto = new UserDTO("John", "Doe", "john.doe@example.com"); String firstName = userDto.firstName(); Note that records are immutable by default, which means that you cannot modify the values of their components once the record is created. If you need to modify the values of a record's components, you can define a custom constructor and/or methods to return modified copies of the record with different values.

User but there is error in userService public List getAllCustomers() { return (userRepository .findAll()) .stream() .map(customerDTOMapper) .collect(Collecters.toList()); } it has error in . stream() .collect(Collecters.toList()); ChatGPT The error is in the spelling of the Collectors class. In the getAllCustomers method, the collect method is spelled as Collecters.toList(), which is causing a compile-time error. The correct spelling is Collectors.toList().

Here's the corrected version of the getAllCustomers method:

scss Copy code public List getAllCustomers() { return userRepository.findAll() .stream() .map(customerDTOMapper::toCustomerDTO) .collect(Collectors.toList()); } Note that I also updated the map method to use a method reference to the toCustomerDTO method of the CustomerDTOMapper class, which is the correct way to use a method reference in this case.

User is customerDTOMapper is error? @Service public class CustomerDTOMapper implements Function<Customer, CustomerDTO> {

@Override
public CustomerDTO apply(Customer customer) {
    // TODO Auto-generated method stub
    return new CustomerDTO(
        customer.getUID(),
        customer.getName(),
        customer.getPhoneNumber(),
        customer.getEmail(),
        customer.getuserType()
    );
}

} ChatGPT No, the CustomerDTOMapper class is not an error. It is a valid implementation of the Function interface that maps a Customer entity to a CustomerDTO.

In the apply method of the CustomerDTOMapper class, the method is using the Customer entity to construct a new CustomerDTO object with the corresponding properties. This is the correct way to map an entity to a DTO using a Function.

However, it's worth noting that the use of a separate CustomerDTOMapper class is optional. In some cases, it may be simpler to define the mapping logic directly in the service or controller method, especially for simpler mappings. For example, you could replace the CustomerDTOMapper class with a lambda expression like this:

less Copy code List customerDTOs = customers .stream() .map(customer -> new CustomerDTO( customer.getUID(), customer.getName(), customer.getPhoneNumber(), customer.getEmail(), customer.getuserType() )) .collect(Collectors.toList()); This lambda expression achieves the same result as the CustomerDTOMapper class, but it's defined directly in the getAllCustomers method instead of being defined as a separate class. Whether to use a separate mapping class or inline the mapping logic depends on the complexity of the mapping and the design of your application.

User if upper code is correct why it is error it makes 'The method stream() is undefined for the type IterableJava(67108964)' ChatGPT The error message "The method stream() is undefined for the type Iterable" indicates that the stream() method cannot be called on an object of type Iterable because the Iterable interface does not define a stream() method.

To fix this error, you can convert the Iterable to a Stream using the StreamSupport class, like this:

scss Copy code public List getAllCustomers() { Iterable customers = userRepository.findAll(); Stream customerStream = StreamSupport.stream(customers.spliterator(), false); return customerStream .map(customerDTOMapper::toCustomerDTO) .collect(Collectors.toList()); } In this code, the customers Iterable is converted to a Stream using the StreamSupport class. The spliterator() method is called on the customers object to obtain a Spliterator, which is then passed to the StreamSupport.stream() method to create a Stream. The second argument to the stream() method is false, which indicates that the stream should not be parallel.

With this modification, the getAllCustomers() method should be able to process the Customer objects returned by the userRepository and map them to CustomerDTO objects using the customerDTOMapper.

User Stream can be changed List ? ChatGPT Yes, you can convert a Stream to a List using the collect() method with Collectors.toList(), like this:

scss Copy code public List getAllCustomers() { Iterable customers = userRepository.findAll(); Stream customerStream = StreamSupport.stream(customers.spliterator(), false); List customerList = customerStream.collect(Collectors.toList()); return customerList .stream() .map(customerDTOMapper::toCustomerDTO) .collect(Collectors.toList()); } In this code, the Stream returned by the userRepository is first converted to a List using the collect() method with Collectors.toList(). The resulting List is then passed to the map() method, which applies the customerDTOMapper to each Customer in the list to create a List.

User good but .map has something error The method map(Function<? super Customer,? extends R>) in the type Stream is not applicable for the arguments (customerDTOMapper::CustomerDTO) ChatGPT The error message "The method map(Function<? super Customer,? extends R>) in the type Stream is not applicable for the arguments (customerDTOMapper::CustomerDTO)" indicates that there is a problem with the lambda expression passed to the map() method. Specifically, it appears that the customerDTOMapper::toCustomerDTO reference is not being interpreted as a valid Function argument.

One possible explanation for this error is that the customerDTOMapper instance is being treated as a regular class rather than as a functional interface. To fix this error, you can modify the customerDTOMapper class to explicitly declare it as a functional interface using the @FunctionalInterface annotation, like this:

java Copy code @FunctionalInterface public interface CustomerDTOMapper { CustomerDTO toCustomerDTO(Customer customer); } With this change, the customerDTOMapper class can be used as a functional interface with a lambda expression or method reference.

Here's an example of how you can modify the getAllCustomers() method to use the customerDTOMapper as a functional interface:

swift Copy code public List getAllCustomers() { Iterable customers = userRepository.findAll(); Stream customerStream = StreamSupport.stream(customers.spliterator(), false); List customerDTOs = customerStream .map(customerDTOMapper::toCustomerDTO) .collect(Collectors.toList()); return customerDTOs; } In this code, the toCustomerDTO() method of the customerDTOMapper instance is used as a method reference with the map() method. The resulting Stream is then collected to a List using the collect() method with Collectors.toList().

User customerDTOMapper is interface? but my code is class @Service public class CustomerDTOMapper implements Function<Customer, CustomerDTO> {

@Override
public CustomerDTO apply(Customer customer) {
    // TODO Auto-generated method stub
    return new CustomerDTO(
        customer.getUID(),
        customer.getName(),
        customer.getPhoneNumber(),
        customer.getEmail(),
        customer.getuserType()
    );
}

} ChatGPT I apologize for my confusion earlier. You are correct that CustomerDTOMapper is a class, not an interface. As a class, it can't be annotated with @FunctionalInterface to indicate that it's a functional interface, because it has a concrete implementation of the apply() method.

The error message "The method map(Function<? super Customer,? extends R>) in the type Stream is not applicable for the arguments (customerDTOMapper::CustomerDTO)" suggests that the map() method is not receiving a valid Function argument. This could be because customerDTOMapper is not being used correctly as a Function instance.

To fix this error, you can modify the map() method call to pass a lambda expression that invokes the apply() method of the customerDTOMapper instance. Here's an example:

swift Copy code public List getAllCustomers() { Iterable customers = userRepository.findAll(); Stream customerStream = StreamSupport.stream(customers.spliterator(), false); List customerDTOs = customerStream .map(customer -> customerDTOMapper.apply(customer)) .collect(Collectors.toList()); return customerDTOs; } In this code, a lambda expression is used to create a Function<Customer, CustomerDTO> instance that invokes the apply() method of the customerDTOMapper instance. This Function instance is then passed as an argument to the map() method. The resulting Stream is then collected to a List using the collect() method with Collectors.toList().

User now ide has no more redline. thx ChatGPT You're welcome! I'm glad to hear that the error has been resolved. If you have any more questions or encounter any other issues, feel free to ask!

choi4624 commented 1 year ago
  1. Lombok 안쓰기
  2. Mapstruct 안쓰기
  3. 자바 요소 자체 혹은 스프링만 쓰고 가능하면 최신 기능 넣기

this mapping is something error, it is defined by 'Reservation' @entity. and Resrvation has Listed orderList @entity @OneToMany(mappedBy = "reservation", cascade = CascadeType.ALL) private List orderList = new ArrayList<>();

Error creating bean with name 'entityManagerFactory' defined in class path resource [org/springframework/boot/autoconfigure/orm/jpa/HibernateJpaConfiguration.class]: Collection 'rest.order.reservation.Model.Reservation.orderList' is 'mappedBy' a property named 'reservation' which does not exist in the target entity 'rest.order.reservation.Model.OrderMenu'

ChatGPT Based on the error message, it seems that there is a mismatch between the mapping definition in the Reservation entity and the corresponding mapping in the OrderMenu entity.

Specifically, the error message indicates that the Reservation entity has a @OneToMany mapping to the OrderMenu entity, using the "reservation" property as the mapped-by attribute. However, the OrderMenu entity does not seem to have a corresponding "reservation" property.

To fix this issue, you may need to ensure that the OrderMenu entity has a "reservation" property that corresponds to the mapped-by attribute in the Reservation entity's @OneToMany mapping. Alternatively, you may need to update the mapped-by attribute in the Reservation entity to match the name of the property that actually maps to the Reservation entity in the OrderMenu entity.

User fix it

and the another error in MenuService Description:

Parameter 0 of constructor in rest.order.reservation.Service.MenuService required a bean of type 'rest.order.reservation.Model.Menu' that could not be found.

The injection point has the following annotations:

Action:

Consider defining a bean of type 'rest.order.reservation.Model.Menu' in your configuration.

how to defining a bean of this code

@Service public class MenuService {

private Menu menuClass;

private MenuRepo menuRepository;

@Autowired
public MenuService(Menu menu, MenuRepo menuRepo) {
    this.menuClass = menu;
    this.menuRepository = menuRepo;
}

public MenuService() {

}
public String menuName() {
    return "WelcomeMenu";
}

public void addMenu(Menu menuClass) {

    menuRepository.save(menuClass); // save가 insert 가 될 수도 있고요 update 가 될 수 있어요 
}

} ChatGPT The error message indicates that Spring is not able to find a bean of type rest.order.reservation.Model.Menu to inject into the constructor of the MenuService class.

To define a bean of this type, you need to add a configuration class or file to your application that specifies how to create the Menu object.

Here's an example of how you could define a Menu bean using Java configuration:

typescript Copy code @Configuration public class MyConfig { @Bean public Menu menu() { return new Menu(); } } This configuration class defines a bean named menu that creates a new Menu object. You can then use this bean in the MenuService class by adding the @Autowired annotation to the menu parameter in the constructor:

typescript Copy code @Service public class MenuService {

private Menu menuClass;

private MenuRepo menuRepository;

@Autowired
public MenuService(Menu menu, MenuRepo menuRepo) {
    this.menuClass = menu;
    this.menuRepository = menuRepo;
}

public MenuService() {

}
public String menuName() {
    return "WelcomeMenu";
}

public void addMenu(Menu menuClass) {

    menuRepository.save(menuClass); // save가 insert 가 될 수도 있고요 update 가 될 수 있어요 
}

} With this configuration, Spring should be able to create a MenuService object with the Menu bean injected into its constructor.

User if menuDTO like below code. how to Make Entity? package rest.order.reservation.Model.DTO;

import java.util.function.Function;

import org.springframework.stereotype.Service;

import rest.order.reservation.Model.Menu;

@Service public class MenuDTOMapper implements Function<Menu, MenuDTO>{

@Override
public MenuDTO apply(Menu menu) {
    // TODO Auto-generated method stub
    return new MenuDTO(
        menu.getName(),
        menu.getIntro(), 
        menu.getPrice(), 
        menu.getType()
    );
}

} ChatGPT To create an Entity for the MenuDTO class, you can define a JPA entity that has fields that match the properties of the MenuDTO class. Here's an example:

typescript Copy code @Entity @Table(name = "menu") public class MenuEntity {

@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;

private String name;

private String intro;

private int price;

private String type;

// constructors, getters, and setters

public MenuEntity() {
}

public MenuEntity(String name, String intro, int price, String type) {
    this.name = name;
    this.intro = intro;
    this.price = price;
    this.type = type;
}

// getters and setters omitted for brevity

} This entity has fields that match the properties of the MenuDTO class, as well as an id field that serves as the primary key for the entity.

To convert between MenuEntity and MenuDTO, you can define a mapper class that implements the Function interface, as shown in your code snippet. Here's an example:

typescript Copy code @Service public class MenuMapper implements Function<MenuEntity, MenuDTO> {

@Override
public MenuDTO apply(MenuEntity menuEntity) {
    return new MenuDTO(
            menuEntity.getName(),
            menuEntity.getIntro(),
            menuEntity.getPrice(),
            menuEntity.getType()
    );
}

} This mapper class converts a MenuEntity object to a MenuDTO object by mapping the fields from the MenuEntity to the corresponding fields in the MenuDTO.

User is this code does not require MenuDTO record? @Service public class MenuMapper implements Function<MenuEntity, MenuDTO> {

@Override
public MenuDTO apply(MenuEntity menuEntity) {
    return new MenuDTO(
            menuEntity.getName(),
            menuEntity.getIntro(),
            menuEntity.getPrice(),
            menuEntity.getType()
    );
}

} ChatGPT The MenuMapper class that you provided does not require a MenuDTO record to be persisted in the database. It is a utility class that maps MenuEntity objects to MenuDTO objects in memory.

The apply method in the MenuMapper class takes a MenuEntity object as input and returns a MenuDTO object that has the same values for its fields. The method simply copies the field values from the MenuEntity to the MenuDTO using the constructor of MenuDTO.

In order to persist data to the database, you would typically use a Repository class that extends JpaRepository. Here's an example of how you could define a MenuRepository class to manage persistence of MenuEntity objects:

java Copy code @Repository public interface MenuRepository extends JpaRepository<MenuEntity, Long> { // add custom methods here if needed } This repository defines CRUD operations for MenuEntity objects, and allows you to persist and retrieve data from the database using Spring Data JPA.

When you want to save a MenuDTO object to the database, you would first convert it to a MenuEntity object using a mapper class like MenuMapper, and then use the MenuRepository to save the entity to the database. Here's an example:

java Copy code @Service public class MenuService {

private final MenuRepository menuRepository;
private final MenuMapper menuMapper;

@Autowired
public MenuService(MenuRepository menuRepository, MenuMapper menuMapper) {
    this.menuRepository = menuRepository;
    this.menuMapper = menuMapper;
}

public MenuDTO saveMenu(MenuDTO menuDTO) {
    MenuEntity menuEntity = new MenuEntity(
            menuDTO.getName(),
            menuDTO.getIntro(),
            menuDTO.getPrice(),
            menuDTO.getType()
    );
    menuRepository.save(menuEntity);
    return menuMapper.apply(menuEntity);
}

// add other service methods here

} In this example, the MenuService class has a MenuRepository field that is used to persist MenuEntity objects to the database. The MenuService also has a MenuMapper field that is used to map MenuEntity objects to MenuDTO objects.

The saveMenu method in the MenuService class takes a MenuDTO object as input, converts it to a MenuEntity object using a constructor, saves the entity to the database using the menuRepository, and then returns a MenuDTO object that is converted from the saved entity using the menuMapper.

User how to convert http Post data to menuDTO 

@PostMapping("/menu/menuaddd")
public String menuadd(Model model) {
    menuService.saveMenu(model);// 메뉴 아이디, 이름, 단가, 등등등
    // 수행하면 menu DB 에 메뉴 등록

    return "menulist";
}

ChatGPT To convert HTTP POST data to a MenuDTO object, you need to extract the data from the HTTP request and create a new MenuDTO object using that data.

Assuming that your HTML form has input fields with name attributes corresponding to the fields in the MenuDTO class (i.e., name, intro, price, and type), you can use the @ModelAttribute annotation to automatically map the form data to a MenuDTO object.

Here's an example of how you could modify the menuadd method to extract the form data and pass it to the menuService:

less Copy code @PostMapping("/menu/menuaddd") public String menuadd(@ModelAttribute MenuDTO menuDTO) { menuService.saveMenu(menuDTO); return "menulist"; } In this example, the @ModelAttribute annotation maps the form data to a new MenuDTO object, which is then passed to the menuService to be saved to the database.

Note that the Model parameter in your original menuadd method is not needed here, since you are not using it to add any data to the model. Also, you need to make sure that the MenuDTO class has a constructor that takes the input data as arguments, or provide a static factory method to create MenuDTO objects from the input data.

User does Entity Menu has public constructor and getter setter. getter and setter is okay? and why it is public?

@Entity @Table(name = "menu") // DB 에 menu 테이블이 만들어져요 public class Menu {

@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "menu_id")
private int mid; // PK 로 mid 등록됨 
@Column(name = "menu_name")
private String name;
@Column(name = "menu_intro")
private String intro;
@Column(name = "price")
private int price;

@Column(name = "type")
@Enumerated(EnumType.STRING)
private MenuType type;

public Menu() {

}

public Menu(Integer mid, String name, String intro, int price, MenuType type) {
    this.mid = mid;
    this.name = name;
    this.intro = intro;
    this.price = price;
    this.type = type;
}

public Menu(Integer name2, String intro2, String price2, int type2) {
}

public int getMid() {
    return mid;
}

public void setMid(int mid) {
    this.mid = mid;
}

public String getName() {
    return name;
}

public void setName(String name) {
    this.name = name;
}

public String getIntro() {
    return intro;
}

public void setIntro(String intro) {
    this.intro = intro;
}

public int getPrice() {
    return price;
}

public void setPrice(int price) {
    this.price = price;
}

public MenuType getType() {
    return type;
}

public void setType(MenuType type) {
    this.type = type;
}

} ChatGPT Yes, the Menu entity has a public constructor and getter/setter methods. The constructor Menu() is a default constructor that takes no arguments, while Menu(Integer mid, String name, String intro, int price, MenuType type) is a constructor that takes in arguments to set the initial values of the instance variables.

The getter methods (getMid(), getName(), getIntro(), getPrice(), getType()) and setter methods (setMid(), setName(), setIntro(), setPrice(), setType()) are public methods used to access and modify the values of the instance variables. These methods are typically generated automatically by an IDE or can be manually written.

The reason why these methods are public is to allow other classes to access and modify the values of the instance variables in a controlled manner. By making the methods public, other classes can interact with the entity object in a way that ensures data integrity and prevents unauthorized access.

User this code has error in menuDTO.getName().. etc 

public MenuDTO saveMenu(MenuDTO menuDTO) {
    Menu menu = new Menu(
            menuDTO.getName(),
            menuDTO.getIntro(),
            menuDTO.getPrice(),
            menuDTO.getType()
    );
    menuRepository.save(menu);
    return menuDTOMapper.apply(menu);
}

The method getName() is undefined for the type MenuDTOJava(67108964) but menuDTO has public 'record' is need to getter or setter in record? ChatGPT No, you do not need to provide getter or setter methods in a record class in Java.

The error message you received is probably caused by a mismatch between the fields of the MenuDTO class and the arguments you pass to its constructor.

Make sure that the constructor of the MenuDTO class takes the same arguments in the same order as the fields in the class definition. For example, if the MenuDTO class has the following definition:

arduino Copy code public record MenuDTO(String name, String intro, int price, MenuType type) {} Then, the constructor call in the saveMenu method should look like this:

less Copy code Menu menu = new Menu( menuDTO.name(), menuDTO.intro(), menuDTO.price(), menuDTO.type() ); Note that the name(), intro(), price(), and type() methods are automatically generated by the Java compiler for record classes, and they return the values of the corresponding fields.

User wait, if my code have MenuDTOMapper it is right way to using upper code?

@Service public class MenuDTOMapper implements Function<Menu, MenuDTO>{

@Override
public MenuDTO apply(Menu menu) {
    // TODO Auto-generated method stub
    return new MenuDTO(
        menu.getName(),
        menu.getIntro(), 
        menu.getPrice(), 
        menu.getType()
    );
}

} ChatGPT Yes, if you have a MenuDTOMapper class that implements the Function<Menu, MenuDTO> interface, you can use it to convert a Menu object to a MenuDTO object.

In your MenuService class, you can use the MenuDTOMapper to convert the MenuDTO object to a Menu object before saving it to the repository:

java Copy code @Service public class MenuService {

private MenuRepo menuRepository; private MenuDTOMapper menuDTOMapper;

@Autowired public MenuService(MenuRepo menuRepo, MenuDTOMapper menuDTOMapper) { this.menuRepository = menuRepo; this.menuDTOMapper = menuDTOMapper; }

public MenuDTO saveMenu(MenuDTO menuDTO) { Menu menu = new Menu( menuDTO.getName(), menuDTO.getIntro(), menuDTO.getPrice(), menuDTO.getType() ); menuRepository.save(menu); return menuDTOMapper.apply(menu); }

} This way, you can convert the MenuDTO to a Menu object using the constructor of the Menu class, and then save it to the repository. After that, you can use the MenuDTOMapper to convert the saved Menu object back to a MenuDTO object and return it.

User code is changed by Menu menu = new Menu( menuDTO.name(), menuDTO.intro(), menuDTO.price(), menuDTO.type() ); why is just typing name(), intro() not using getName().. ChatGPT In Java 16 and later, record types were introduced, which is a new way to declare classes that are mostly intended to store data. Records provide concise syntax for declaring such classes by eliminating the need for explicitly declaring fields, constructors, and basic methods like getters and setters.

When you define a record type in Java, the compiler automatically generates a constructor that initializes the record's components and accessor methods with the same name as the component.

So in your code, menuDTO is a record type, and it has four components: name, intro, price, and type. When you call menuDTO.name(), it returns the value of the name component, and when you call menuDTO.intro(), it returns the value of the intro component, and so on.

Therefore, you don't need to use getName(), getIntro(), getPrice(), or getType() methods, as the generated accessor methods are equivalent and have the same functionality.

choi4624 commented 1 year ago

2023-05-04_020651 2023-05-04_020758