如何通过传递两个参数来修复方法调用函数

Question

我正在创建一个带有登录页面的待办事项列表，其中有一个两个输入表作为名称，名称中的事物，事物不向服务器发送数据这是存储库...public interface TodosRepository extends CrudRepository<Todos, Long> {    Todos findByName(String name,String things);    List<Todos> findByCompletedAndUserId(boolean complated, Long Id);}这是我的实体@Entitypublic class Todos {    @Id    @GeneratedValue(strategy = GenerationType.IDENTITY)    private Long id;    private String things;    private String name;    private boolean completed;    @ManyToOne(fetch = FetchType.LAZY)    @JoinColumn(name = "user_id")    private User user;    public Todos() {    }    public Todos(String name,String things, boolean completed) {        this.things = things;        this.name = name;        this.completed = completed;    }    public String getThings() {        return things;    }    public void setThings(String things) {        this.things = things;    }    public Long getId() {        return id;    }    public void setId(Long id) {        this.id = id;    }    public String getName() {        return name;    }    public void setName(String name) {        this.name = name;    }    public boolean isCompleted() {        return completed;    }    public void setCompleted(boolean completed) {        this.completed = completed;    }    public User getUser() {        return user;    }    public void setUser(User user) {        this.user = user;    }    @Override    public String toString() {        return "Todos [id=" + id + ", name=" + name + ",things=" + things +", completed=" + completed + "]";    }}java.util.NoSuchElementException：null

Answer 1

我认为拼写错误，这就是原因，completed而不是complated List<Todos> findByCompletedAndUserId(boolean complated, Long Id); 参数名称应该与方法名称相同。就像方法名称有userId而参数只有id




Answer 2

你没有这个注释，请尝试使用下面的这一行@Entity

@Table(name = "your-table")