java - Spring Boot“无法执行 CommandLineRunner 错误”
问题描述
我尝试将一对多注释添加到我的 Spring Boot 项目中,但是当我运行我的项目时,我收到“无法执行 CommandLineRunner”错误。我希望用户表中的用户拥有多个城市。所以,我尝试添加 OneToMany 注释。
用户类
package io.javabrains.springsecurity.jpa.models;
import com.spring.weather.*;
import java.util.*;
import java.util.List;
import javax.persistence.CascadeType;
import javax.persistence.Entity;
import javax.persistence.FetchType;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.JoinTable;
import javax.persistence.OneToMany;
import javax.persistence.Table;
@Entity
@Table(name="app_user")
public class User {
@Id
@GeneratedValue(strategy =GenerationType.AUTO)
private int id;
private String userName;
private String password;
private boolean active;
private String role;
private String city;
@OneToMany(targetEntity = UserCity.class,cascade = CascadeType.ALL)
@JoinTable(name="USER_CITY",joinColumns=@JoinColumn(name="m_user_id"),
inverseJoinColumns=@JoinColumn(name="cityId"))
private List<UserCity> usercity;
public User() {
super();
// TODO Auto-generated constructor stub
}
public List<UserCity> getUsercity() {
return usercity;
}
public void setUsercity(List<UserCity> usercity) {
this.usercity = usercity;
}
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
public String getUserName() {
return userName;
}
public void setUserName(String userName) {
this.userName = userName;
}
public String getPassword() {
return password;
}
public void setPassword(String password) {
this.password = password;
}
public boolean isActive() {
return active;
}
public void setActive(boolean active) {
this.active = active;
}
public String getRole() {
return role;
}
public void setRole(String role) {
this.role = role;
}
public String getCity() {
return city;
}
public void setCity(String city) {
this.city = city;
}
}
用户城市等级
package io.javabrains.springsecurity.jpa.models;
import javax.persistence.Entity;
import javax.persistence.FetchType;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.ManyToOne;
import javax.persistence.Table;
@Entity
@Table(name="user_city")
public class UserCity {
@Id
@GeneratedValue(strategy =GenerationType.AUTO)
private int cityId;
private String cityName;
public UserCity() {
super();
// TODO Auto-generated constructor stub
}
public UserCity(int cityId, String cityName, User mUser) {
super();
this.cityId = cityId;
this.cityName = cityName;
this.mUser = mUser;
}
@ManyToOne
private User mUser;
public int getCityId() {
return cityId;
}
public void setCityId(int cityId) {
this.cityId = cityId;
}
public String getCityName() {
return cityName;
}
public void setCityName(String cityName) {
this.cityName = cityName;
}
public User getmUser() {
return mUser;
}
public void setmUser(User mUser) {
this.mUser = mUser;
}
}
用户存储库
package io.javabrains.springsecurity.jpa;
import java.util.Optional;
import org.springframework.data.jpa.repository.JpaRepository;
import org.springframework.stereotype.Repository;
import io.javabrains.springsecurity.jpa.models.User;
public interface UserRepository extends JpaRepository<User, Integer> {
Optional<User> findByUserName(String userName);
}
用户城市资料库
package io.javabrains.springsecurity.jpa;
import org.springframework.data.jpa.repository.JpaRepository;
import io.javabrains.springsecurity.jpa.models.User;
import io.javabrains.springsecurity.jpa.models.UserCity;
public interface CityRepository extends JpaRepository<UserCity,id>{
}
Spring 应用类
package io.javabrains.springsecurity.jpa;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.boot.CommandLineRunner;
import org.springframework.boot.SpringApplication;
import org.springframework.boot.autoconfigure.SpringBootApplication;
import org.springframework.context.annotation.Bean;
import org.springframework.data.jpa.repository.config.EnableJpaRepositories;
import org.springframework.security.crypto.bcrypt.BCryptPasswordEncoder;
import com.spring.weather.WeatherService;
import io.javabrains.springsecurity.jpa.models.User;
import io.javabrains.springsecurity.jpa.models.UserCity;
@SpringBootApplication
@EnableJpaRepositories(basePackageClasses = UserRepository.class)
public class SpringsecurityApplication implements CommandLineRunner{
@Bean
public WeatherService ws() {
return new WeatherService ();
}
@Autowired
UserRepository userRepository;
CityRepository cityRepository;
public static void main(String[] args) {
SpringApplication.run(SpringsecurityApplication.class, args);
}
@Override
public void run(String... args) throws Exception {
// TODO Auto-generated method stub
System.out.println("Application Running.");
User adminUser= new User();
UserCity ucity=new UserCity();
UserCity ucity2=new UserCity();
ucity.setCityName("amsterdam");
adminUser.setUserName("Admin");
adminUser.setPassword(new BCryptPasswordEncoder().encode("pass"));
adminUser.setRole("ROLE_ADMIN");
adminUser.setActive(true);
adminUser.setCity("bologna");
ucity.setmUser(adminUser);
userRepository.save(adminUser);
cityRepository.save(ucity);
User newUser= new User();
newUser.setUserName("User");
newUser.setPassword(new BCryptPasswordEncoder().encode("pass"));
newUser.setRole("ROLE_USER");
newUser.setActive(true);
newUser.setCity("maribor");
ucity2.setmUser(newUser);
userRepository.save(newUser);
cityRepository.save(ucity2);
}
}
解决方案
cityRepository
您遇到的问题,更具体地说是主应用程序第 54 行的 NullPointerException,是由未实例化的事实引起的。
查看您的配置,我看到您只UserRepository
使用@EnableJpaRepositories
注释注册了。
尝试将 也添加CityRepository
到@EnableJpaRepositories
,并将此 bean 指定为自动装配的候选对象(也添加@Autowired
到此 bean,就像您为 所做的那样UserRepository
)
对于一个好的实践,遵循 MVC 结构,最好是你所有的 spring 存储库,负责与数据库的所有 CRUD 操作的 bean 在同一个包下
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