Question

我正在使用Spring Boot + JPA保存某些实体的应用程序。但是，这些实体（当然还有Spring配置）是应用程序中唯一依赖Spring / JPA的东西。这就是为什么我们想将应用程序分为两个部分。应该独立于Spring和任何JPA相关事物的核心应用程序以及Spring应用程序。为此，我们在核心应用程序中创建了实体的接口和基础实现，在Spring应用程序中创建了派生实现，该实现仅覆盖getter并使用属性访问。但是现在Spring抱怨说这些字段不存在。

这是核心应用程序中的类，基本上只是一堆getter和setter：

public class UserImpl implements User {

protected Set<UserProfile> userProfiles = new HashSet<>();

protected Set<Playlist> playlists = new HashSet<>();

protected String userName;
protected String password;
protected String securityQuestion;
protected String securityQuestionAnswer;
protected String token;

public UserImpl() {
}

public UserImpl(String userName) {
    this.userName = userName;
}

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 String getSecurityQuestion() {
    return securityQuestion;
}

public void setSecurityQuestion(String securityQuestion) {
    this.securityQuestion = securityQuestion;
}

public String getSecurityQuestionAnswer() {
    return securityQuestionAnswer;
}

public void setSecurityQuestionAnswer(String securityQuestionAnswer) {
    this.securityQuestionAnswer = securityQuestionAnswer;
}

public void setUserProfiles(Set<UserProfile> userProfiles) {
    this.userProfiles = userProfiles;
}

public List<UserProfile> getUserProfiles() {
    return new ArrayList<>(userProfiles);
}

public void setToken(String token) {
    this.token = token;
}

public String getToken() {
    return token;
}

public void setPlaylists(Set<Playlist> playlists) {
    this.playlists = playlists;
}

public Set<Playlist> getPlaylists() {
    return playlists;
}

}

这是派生类：

@Entity
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
@Table(name = "users", schema = "musictinder")
public class JPAUserImpl extends UserImpl {

@Id @GeneratedValue private long id;

public long getId() {
    return id;
}

public void setId(long id) {
    this.id = id;
}

@Override
@Column(name = "user_name")
@AccessType(AccessType.Type.PROPERTY)
public String getUserName() {
    return super.getUserName();
}

@Override
@Column(name = "password")
@AccessType(AccessType.Type.PROPERTY)
public String getPassword() {
    return super.getPassword();
}

@Override
@Column(name = "security_question")
@AccessType(AccessType.Type.PROPERTY)
public String getSecurityQuestion() {
    return super.getSecurityQuestion();
}

@Override
@Column(name = "security_question_answer")
@AccessType(AccessType.Type.PROPERTY)
public String getSecurityQuestionAnswer() {
    return super.getSecurityQuestionAnswer();
}

@Override
@Column(name = "token")
@AccessType(AccessType.Type.PROPERTY)
public String getToken() {
    return super.getToken();
}

@Override
@ElementCollection
@OneToMany(cascade = CascadeType.ALL)
@AccessType(AccessType.Type.PROPERTY)
public Set<Playlist> getPlaylists() {
    return super.getPlaylists();
}

@Override
@ElementCollection
@OneToMany(cascade = CascadeType.ALL)
@AccessType(AccessType.Type.PROPERTY)
public List<UserProfile> getUserProfiles() {
    return super.getUserProfiles();
}
}

更新：我发现Spring只抱怨userName字段。当我将其添加到派生类时，应用程序不会崩溃。由于某种原因，它仅与其他字段一起使用。现在我实际上比以前更加困惑。

Answer 1

您可能需要在超级类UserImpl上使用@MappedSuperclass批注

请参阅文档https://docs.oracle.com/javaee/5/api/javax/persistence/MappedSuperclass.html

Spring Boot + JPA：派生类中的“无法找到具有给定名称的属性”

1 个答案: