Java 没有外键的JPA OneToOne关联总是失败

我是JPA/Hibernate新手， 我正在创建一个简单的连接，以获取不同表中的一些数据 让我们这样看：

其目的是基于M_USER.username列获取M_配置文件数据，但您可以清楚地看到，M_配置文件表中没有外键

我只是尝试使用下面的代码，但没有结果，总是得到错误

用户实体类

@Entity
@Table(name = "M_USER")
public class User {

        @Id
        @Column(name = "id")
        private String uuid;

        @Column(name = "email")
        private String email;

        @Column(name = "username")
        private String username;

        @OneToOne(fetch = FetchType.LAZY)
        @MapsId("username")
        private Profile profile;

        public String getId() {
            return id;
        }
        public void setId(String id) {
            this.id = id;
        }

        public String getEmail() {
            return email;
        }
        public void setEmail(String email) {
            this.email = email;
        }

        public String getUsername() {
            return username;
        }
        public void setUsername(String username) {
            this.username = username;
        }

        public Profile getProfile() {
            return profile;
        }
        public void setProfile(Profile profile) {
            this.profile = profile;
        }
    }

@Entity
@Table(name = "M_PROFILE")
public class Profile {

    private String username;
    private String phone;
    private String address;

    @Id
    @Column(name = "username", nullable = false)
    public String getUsername() {
        return username;
    }

    public void setUsername(String username) {
        this.username = username;
    }

    @Basic
    @Column(name = "phone")
    public String getPhone() {
        return phone;
    }

    public void setPhone(String phone) {
        this.phone = phone;
    }

    @Basic
    @Column(name = "address")
    public String getAddress() {
        return address;
    }

    public void setAddress(String address) {
        this.address = address;
    }
}

配置文件实体类

@Entity
@Table(name = "M_USER")
public class User {

        @Id
        @Column(name = "id")
        private String uuid;

        @Column(name = "email")
        private String email;

        @Column(name = "username")
        private String username;

        @OneToOne(fetch = FetchType.LAZY)
        @MapsId("username")
        private Profile profile;

        public String getId() {
            return id;
        }
        public void setId(String id) {
            this.id = id;
        }

        public String getEmail() {
            return email;
        }
        public void setEmail(String email) {
            this.email = email;
        }

        public String getUsername() {
            return username;
        }
        public void setUsername(String username) {
            this.username = username;
        }

        public Profile getProfile() {
            return profile;
        }
        public void setProfile(Profile profile) {
            this.profile = profile;
        }
    }

@Entity
@Table(name = "M_PROFILE")
public class Profile {

    private String username;
    private String phone;
    private String address;

    @Id
    @Column(name = "username", nullable = false)
    public String getUsername() {
        return username;
    }

    public void setUsername(String username) {
        this.username = username;
    }

    @Basic
    @Column(name = "phone")
    public String getPhone() {
        return phone;
    }

    public void setPhone(String phone) {
        this.phone = phone;
    }

    @Basic
    @Column(name = "address")
    public String getAddress() {
        return address;
    }

    public void setAddress(String address) {
        this.address = address;
    }
}

我打电话时遇到了不同的错误

User user = userRepository.findByUsername("aswzen");
String phone = user.getProfile().getPhone();

比如这个

“USER0”“PROFILE\u USERNAME”：无效标识符

需要帮助，提前谢谢

---

注：我没有权利更改这张桌子

尝试指定联接列：

@OneToOne
@JoinColumn(name = "username", referencedColumnName = "username")
private profile profile;

---

此处不需要

@MapsId

。

尝试指定联接列：

@OneToOne
@JoinColumn(name = "username", referencedColumnName = "username")
private profile profile;

---

您不需要在此处使用

@MapsId

。

我建议为您的表创建第三个具有关系的表

@Entity
@Table(name = "M_USER")
public class User {

@OneToOne(fetch = FetchType.LAZY)
@JoinTable(
name = "user_profile",
joinColumns = @JoinColumn(name="user_id"),
inverseJoinColumns = @JoinColumn(name="profile_username")
 )
private Profile profile;


UPDATE user_profile 
SET user_id = (
    SELECT id
    FROM M_USER
);


UPDATE user_profile 
SET profile_username = (
    SELECT username
    FROM M_USER
);

---

我建议为您的表创建具有关系的第三个表

@Entity
@Table(name = "M_USER")
public class User {

@OneToOne(fetch = FetchType.LAZY)
@JoinTable(
name = "user_profile",
joinColumns = @JoinColumn(name="user_id"),
inverseJoinColumns = @JoinColumn(name="profile_username")
 )
private Profile profile;


UPDATE user_profile 
SET user_id = (
    SELECT id
    FROM M_USER
);


UPDATE user_profile 
SET profile_username = (
    SELECT username
    FROM M_USER
);

---

我不确定，但可能有必要将注释添加到getter中。它肯定会起作用。。代码已在当前环境Private profile上运行；-->使其成为资本坑打字错误..我的已经是私有配置文件。。我不确定，但可能有必要将注释添加到getter中。它肯定会起作用。。代码已在当前环境Private profile上运行；-->使其成为资本坑打字错误..我的已经是私有配置文件。。无法工作得到错误..从users user0_u0中选择user0_0.profile作为profile1_14_u14等，其中user0_0.username=？；错误消息=ORA-00904:“M_USER0_uu”。“PROFILE”：无效标识未工作，出现错误..从用户USER0_u选择USER0_u.PROFILE作为profile1_14_uu等，其中USER0_u.username=？；错误Msg=ORA-00904:“M_用户0_”。“配置文件”：无效标识我无权更改或创建表。。因此，与isi一样，该表也无权更改或创建该表。。所以这张桌子还是原样