Question

我已使用GSON将JSON转换为POJO。我正在使用JPA的save（）方法将Employee实体对象存储到mysql中。但是我收到一个错误消息，提示“无法确定地址的类型”。那我该怎么办呢？

这是错误：无法确定：地址的类型

Employee.java

package com.example.demo;

import java.math.BigDecimal;
import java.util.Map;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.Id;

import com.google.gson.annotations.Expose;

@Entity
public class Employee {

@Id
private int id;
private String name;
private int age;
private BigDecimal salary;
private String designation;
private Address address;
private long[] phoneNumbers;

/*Getter and Setter Methods*/
public int getId() {
    return id;
}

public void setId(int id) {
    this.id = id;
}

public String getName() {
    return name;
}

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

public int getAge() {
    return age;
}

public void setAge(int age) {
    this.age = age;
}

public BigDecimal getSalary() {
    return salary;
}

public void setSalary(BigDecimal salary) {
    this.salary = salary;
}

public String getDesignation() {
    return designation;
}

public void setDesignation(String designation) {
    this.designation = designation;
}


public long[] getPhoneNumbers() {
    return phoneNumbers;
}

public void setPhoneNumbers(long[] phoneNumbers) {
    this.phoneNumbers = phoneNumbers;
}

public Address getAddress() {
    return address;
}

public void setAddress(Address address) {
    this.address = address;
}


}

Address.java

package com.example.demo;

import javax.persistence.Entity;
import javax.persistence.Id;

//@Entity
public class Address {
@Id
private String street;
private String city;
private int zipCode;
public String getStreet() {
    return street;
}
public void setStreet(String street) {
    this.street = street;
}
public String getCity() {
    return city;
}
public void setCity(String city) {
    this.city = city;
}
public int getZipCode() {
    return zipCode;
}
public void setZipcode(int zipcode) {
    this.zipCode = zipcode;
}
@Override
public String toString(){
    return getStreet() + ", "+getCity()+", "+getZipCode();
}
}

控制器类

package com.example.demo;

import net.sf.json.*;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.GetMapping;
import org.springframework.web.bind.annotation.RequestBody;
import org.springframework.web.bind.annotation.ResponseBody;

import com.google.gson.Gson;
import com.google.gson.GsonBuilder;
import com.google.gson.JsonObject;

import net.sf.json.JSONObject;

@Controller    // This means that this class is a Controller
@RequestMapping(path="/demo") // This means URL's start with /demo 
(after Application path)
public class MainController {
@Autowired // This means to get the bean called userRepository
           // Which is auto-generated by Spring, we will use it 
to handle the data
private UserRepository userRepository;

@GetMapping(path="/add") // Map ONLY GET Requests
public @ResponseBody String addNewUser (@RequestBody String json) { 
    // @ResponseBody means the returned String is the response, not a view name
    // @RequestParam means it is a parameter from the GET or POST request


    //JSONObject jsonObject = JSONObject.fromObject(json);
    Gson gson=new GsonBuilder().create();
    Employee employee =gson.fromJson(json,Employee.class);
    userRepository.save(employee);
    return "Successfully added to database using JPA!";
    }

@GetMapping(path="/all")
public @ResponseBody Iterable<Employee> getAllUsers() {
    // This returns a JSON or XML with the users
    return userRepository.findAll();
}
}

Answer 1

尝试为您的实体类添加可序列化的实现。即。

@Entity
public class Employee implements Serializable{
}

Answer 2

每当在类中使用@Entity注释并尝试将其实例保存在数据库中时。最初，在休眠状态下会创建一个create table命令，该命令会在数据库中使用给定的规范创建表。与在employee表中一样，您可以指定所有数据成员的数据类型。当光标进入“地址”时，它给出了错误，因为休眠无法找到与此相关的任何数据类型或与此相关的任何表。正如您在Address类中评论@Entity注释。因此，不会在数据库中创建与此有关的表。地址是此休眠需要的类（表）的类引用。由于class（table）不存在，因此出现错误。

如何使用jpa在mysql中存储Java对象？

2 个答案: