这个问题在这里已经有答案了:
为什么我的spring@autowired字段为空(22个答案)
12天前关门了。
所有人。我试图建立一个游戏应用程序,但遇到了一个问题。我创建了一个spring存储库、实体和服务,但每当调用后者时,我都会遇到一个异常,说明:
Exception in thread "main" java.lang.NullPointerException: Cannot invoke "com.example.demo.repo.PlayerRepository.findById(Object)" because "this.playerRepository" is null
at com.example.demo.service.PlayerService.findPlayerById(PlayerService.java:16)
at com.example.demo.DemoApplication.main(DemoApplication.java:16)
以下是我提供的课程,展示了我在课程中包括的内容:
游戏性:
package com.example.demo.model;
import javax.persistence.*;
import java.util.Objects;
@Entity
@Table(name = "player", schema = "public", catalog = "gamestats")
public class PlayerEntity {
private long id;
private int health;
private int damage;
private int absorb;
private int regen;
private int fire;
@Id
@Column(name = "id")
public long getId() {
return id;
}
public void setId(long id) {
this.id = id;
}
@Basic
@Column(name = "health")
public int getHealth() {
return health;
}
public void setHealth(int health) {
this.health = health;
}
@Basic
@Column(name = "damage")
public int getDamage() {return this.damage; }
public void setDamage(int damage) {
this.damage = damage;
}
@Basic
@Column(name = "absorb")
public int getAbsorb() {
return absorb;
}
public void setAbsorb(int absorb) {
this.absorb = absorb;
}
@Basic
@Column(name = "regen")
public int getRegen() {
return regen;
}
public void setRegen(int regen) {
this.regen = regen;
}
@Basic
@Column(name = "fire")
public int getFire() {
return fire;
}
public void setFire(int fire) {
this.fire = fire;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
PlayerEntity that = (PlayerEntity) o;
return id == that.id && health == that.health && damage == that.damage && absorb == that.absorb && regen == that.regen && fire == that.fire;
}
@Override
public int hashCode() {
return Objects.hash(id, health, damage, absorb, regen, fire);
}
}
播放器存储库:
package com.example.demo.repo;
import com.example.demo.model.PlayerEntity;
import org.springframework.data.jpa.repository.JpaRepository;
import org.springframework.stereotype.Repository;
@Repository
public interface PlayerRepository extends JpaRepository<PlayerEntity, Long> {
}
播放器服务:
package com.example.demo.service;
import com.example.demo.model.PlayerEntity;
import com.example.demo.repo.PlayerRepository;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.stereotype.Service;
@Service
public class PlayerService{
@Autowired
PlayerRepository playerRepository;
private PlayerEntity playerEntity;
public PlayerEntity findPlayerById(long id) {
return playerRepository.findById(id).get();
}
public void safePlayer(PlayerEntity test){ playerRepository.save(test); }
public int getPlayerStrength(long id) {
return playerRepository.findById(id).get().getHealth();
}
}
演示应用程序:
package com.example.demo;
import com.example.demo.service.PlayerService;
import org.json.JSONException;
import org.springframework.boot.SpringApplication;
import org.springframework.boot.autoconfigure.SpringBootApplication;
import java.io.IOException;
@SpringBootApplication
public class DemoApplication {
public static void main(String[] args) throws IOException, InterruptedException, JSONException {
SpringApplication.run(DemoApplication.class, args);
PlayerService playerService = new PlayerService();
playerService.findPlayerById(1);
}
我真的很感激你的帮助,因为我真的被卡住了。
2条答案
按热度按时间hjzp0vay1#
服务需要自动连接存储库;只有在由spring上下文创建时才会发生这种情况。通过在main via中创建
new
,Spring没有机会向它注入任何东西,所以它的能量场仍然存在null
.如果您想使用服务,您需要从一个方法访问它,以确保上下文已经正确创建。你可以通过使用
@PostConstruct
方法。ffvjumwh2#
您必须使用optional来确定是否有一个空值对应于任何玩家id。您可以遵循以下语法。