java - 我不能在我的代码中使用 findOne() 方法

Question

我的应用程序有错误，因为我使用了 findOne() 方法。在我的简单代码下面。在 User 类中，我的 id 是 String email ，这是我试图在我的 UserService 类中使用的 id ，如下所示:

public User findUser(String email){
    return userRepository.findOne(email);
}

但是我有这个错误:

method findOne in interface org.springframework.data.repository.query.QueryByExampleExecutor cannot be applied to given types;
required: org.springframework.data.domain.Example
found: java.lang.String
reason: cannot infer type-variable(s) S (argument mismatch; java.lang.String cannot be converted to org.springframework.data.domain.Example)

用户类:

@Entity
@Data
@Table(name = "User")
public class User {
    @Id
    @Email
    @NotEmpty
    @Column(unique = true)
    private String email;

    @NotEmpty
    private String name;

    @NotEmpty
    @Size(min = 5)
    private String password;

    @OneToMany(mappedBy = "user", cascade = CascadeType.ALL)
    private List<Task> tasks;

    @ManyToMany(cascade = CascadeType.ALL)
    @JoinTable(name = "USER_ROLE", joinColumns = {
        @JoinColumn(name = "USER_EMAIL", referencedColumnName = "email")
    }, inverseJoinColumns = {@JoinColumn(name = "ROLE_NAME", referencedColumnName = "name")})
    private List<Role> roles;
}

和 UserRepository:

public interface UserRepository extends JpaRepository<User, String> {
}

Answer 1

使用findById或 getOne而不是 findOne当您只想按 ID 搜索时。

public User findUser(String email){
    return userRepository.getOne(email); // throws when not found or
                                         // eventually when accessing one of its properties
                                         // depending on the JPA implementation
}

public User findUser(String email){
    Optional<User> optUser = userRepository.findById(email); // returns java8 optional
    if (optUser.isPresent()) {
        return optUser.get();
    } else {
        // handle not found, return null or throw
    }
}

函数findOne()收到 Example<S> ，该方法用于实例查找，需要提供实例对象和需要查询的字段。

您可以通过示例了解如何使用查找。

https://docs.spring.io/spring-data/jpa/docs/current/reference/html/#query-by-example.matchers

但基本上是这样的。

User user = new User();                          
person.setName("Dave");                           

ExampleMatcher matcher = ExampleMatcher.matching()     
    .withIgnorePaths("name")                         
    .withIncludeNullValues()                             
    .withStringMatcherEnding();

Example<User> example = Example.of(user, matcher);