Question

我正在通过Hibernate运行查询，并且正在返回列表。但是列表中包含对象而不是我的实体...我不是Hibernate的专业人士，但是我的其他查询运行良好。我不知道我现在的错误是什么。

List getAllUsersBasedOnMandant(Long id) {
List users = null;
try {
  startOperation(false);
  Query query = getSession().createQuery(" from UserEntity where mandant = '" + id + "'");
  users = query.list();
} catch (HibernateException e) {
  handleException(e);
} finally {
  getSession().close();
}
return users;

}

我的输出如下：

[int_plan.entity.UserEntity @ 4b82b237，int_plan.entity.UserEntity @ 7141a0bb，int_plan.entity.UserEntity @ 65b0a12c]

我的实体看起来像这样：

@Entity
@Table(name = "user", schema = "entw_pares")
public class UserEntity {
@Expose() private Long userId;
@Expose() private String gender;
@Expose() private String firstname;
@Expose() private String lastname;
@Expose() private String username;
@Expose() private String email;
private String password;
private String secQuestion;
private String secAnswer;
private String saltAnswer;
private String salt;
private String emailValidationCode;
private Long expireTime;
private Boolean emailEnable = false;
@Expose() private Timestamp dateCreated;
@Expose() private Timestamp dateUpdated;
private Boolean admin = false;
private MandantEntity mandantEntity;


@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "user_id", nullable = false)
public Long getUserId() {
    return userId;
}

public void setUserId(Long userId) {
    this.userId = userId;
}


@Basic
@Column(name = "gender", nullable = false)
public String getGender() { return gender; }

public void setGender(String gender){
    this.gender = gender;
}


@Basic
@Column(name = "firstname", nullable = false)
public String getFirstname() { return firstname; }

public void setFirstname(String firstnme) { this.firstname = firstnme; }


@Basic
@Column(name = "lastname", nullable = false)
public String getLastname() { return lastname; }

public void setLastname(String lastname) { this.lastname = lastname; }


@Basic
@Column(name = "username")
public String getUsername() {
    return username;
}

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


@Basic
@Column(name = "email", nullable = false)
public String getEmail() {
    return email;
}

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

@Basic
@Column(name = "password", nullable = false)
public String getPassword() {
    return password;
}

public void setPassword(String password) {
    this.password = password;
}

@Basic
@Column(name = "sec_question")
public String getSecQuestion() { return secQuestion; }

public void setSecQuestion(String sec_question) { this.secQuestion = 
sec_question; }

@Basic
@Column(name = "sec_answer")
public String getSecAnswer() { return secAnswer; }

public void setSecAnswer(String secAnswer) { this.secAnswer = secAnswer; }

@Basic
@Column(name = "salt_answer")
public String getSaltAnswer() { return saltAnswer; }

public void setSaltAnswer(String saltAnswer) { this.saltAnswer = saltAnswer; 
}

@Basic
@Column(name = "salt")
public String getSalt() {
    return salt;
}

public void setSalt(String salt) {
    this.salt = salt;
}

@Basic
@Column(name = "email_validation_code")
public String getEmailValidationCode() {
    return emailValidationCode;
}

public void setEmailValidationCode(String emailValidationCode) {
    this.emailValidationCode = emailValidationCode;
}

@Basic
@Column(name = "expire_time", nullable = false)
public Long getExpireTime() {
    return expireTime;
}

public void setExpireTime(Long expireTime) {
    this.expireTime = expireTime;
}

@Basic
@Column(name = "email_enable")
public Boolean getEmailEnable() {
    return emailEnable;
}

public void setEmailEnable(Boolean emailEnable) {
    this.emailEnable = emailEnable;
}

@Basic
@CreationTimestamp
@Column(name = "date_created")
public Timestamp getDateCreated() {
    return dateCreated;
}

public void setDateCreated(Timestamp dateCreated) {
    this.dateCreated = dateCreated;
}

@Basic
@UpdateTimestamp
@Column(name = "date_updated")
public Timestamp getDateUpdated() {
    return dateUpdated;
}

public void setDateUpdated(Timestamp dateUpdated) {
    this.dateUpdated = dateUpdated;
}

@Basic
@Column(name = "admin")
public Boolean getAdmin() {
    return admin;
}

public void setAdmin(Boolean admin) {
    this.admin = admin;
}

public void applyValue(Field field, Object value) throws 
IllegalAccessException {
    field.set(this, value);
}


@Override
public boolean equals(Object o) {
    if (this == o) return true;
    if (o == null || getClass() != o.getClass()) return false;
    UserEntity that = (UserEntity) o;
    return userId == that.userId &&
            Objects.equals(gender, that.gender)&&
            Objects.equals(firstname,that.firstname)&&
            Objects.equals(lastname,that.lastname)&&
            Objects.equals(username, that.username) &&
            Objects.equals(email, that.email) &&
            Objects.equals(password, that.password) &&
            Objects.equals(secQuestion,that.secQuestion) &&
            Objects.equals(secAnswer, that.secAnswer) &&
            Objects.equals(salt, that.salt) &&
            Objects.equals(emailValidationCode, that.emailValidationCode) &&
            Objects.equals(emailEnable, that.emailEnable) &&
            Objects.equals(dateCreated, that.dateCreated) &&
            Objects.equals(dateUpdated, that.dateUpdated) &&
            Objects.equals(admin, that.admin);
}

@Override
public int hashCode() {
    return Objects.hash(userId, gender, firstname, lastname, username, email, 
password, secQuestion, secAnswer,
            salt,
            emailValidationCode,
            emailEnable,
            dateCreated, dateUpdated, admin);
}

@ManyToOne
@JoinColumn(name = "mandant_id", referencedColumnName = "mandant_id")
public MandantEntity getMandant() {
    return mandantEntity;
}

public void setMandant(MandantEntity mandant) {
    this.mandantEntity = mandant;
}

有什么想法我做错了吗？

Answer 1

我希望我正确理解了这个问题，希望我的回答对您有所帮助。

请使用query.setParameter([name of parameter], [value])，它将帮助您进行缩略。该链接的更多详细信息：https://www.mkyong.com/hibernate/hibernate-parameter-binding-examples/
您的实体现在是users列表中的对象。您可以将列表定义为所需的对象类型，例如List<UserEntity> users = q.getResultList()

2.a现在，您可以处理列表中的单个元素，例如
```
if(!users.isEmpty()) {
    for(UserEntity uEntity : users){
        uEntity.doSomething()
    }
    return users.get(0)}
```

我希望这对您有帮助

Answer 2

我看了一会儿，我看到这个问题的答案可能与您的问题有关。

最终，这就是您要寻找的答案：

Query.list（）返回原始List对象，而不是通用List。

您应该只能将其强制转换为List，并忽略不安全的内容   发出警告。

我认为您混淆了将原始列表与   List元素的实际运行时类型为Object。只是因为   List.get（i）的编译时签名返回一个Object   表示返回的元素的实际类型是Object at   运行时。

贷记给用户“ matt b”，原始答案：https://stackoverflow.com/a/5913021/9891058

HibernateQuery返回对象列表而不是实体

2 个答案: