首页 > 解决方案 > 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。

您可以在附件中看到错误。

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);
        

    }

}

标签: javaspring-bootspring-data-jpaone-to-many

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@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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