java - 如何使用 JPA 和 Hibernate 通过自定义对象实体属性进行查询

Question

编写一个 DAO 方法以通过电子邮件检索 appUsers，其中电子邮件属于 EmailAddress 类。一直在寻找如何做到这一点无济于事。有没有办法根据自定义对象的值检索记录？当我有一个 String 或 int 类型的属性时，这似乎有效，但我如何获得与我的 EmailAddress 对象匹配的记录？

尝试了以下但不工作:

public List<AppUser> findByEmail(EmailAddress email) {
    log.debug("finding " + getTable() + " instance by example");

    AppUser appUser = new AppUser();
    appUser.setEmail(email);

    try {
        List<AppUser> results = this.sessionFactory.getCurrentSession()
                .createCriteria(AppUser.class)
                .add(Example.create(appUser)).list();
        log.debug("find by example successful, result size: "
                + results.size());
        return results;
    } catch (RuntimeException re) {
        log.error("find by example failed", re);
        throw re;
    }
}

这是我的 AppUser 类的相关部分

@Entity
@Table(name="AppUser")
public class AppUser extends BaseEntity {
    ...

    @Column(name="Email")
    private EmailAddress email;

    ...

    /**
     * Gets this AppUser's associated email
     * 
     * @return this AppUser's associated email
     */
    public EmailAddress getEmail() {
        return this.email;
    }

    /**
     * Sets this AppUser's associated email
     * 
     * @param email this AppUser's associated email
     */
    public void setEmail(EmailAddress email) {
        this.email = email;
    }   
}

这是我的 EmailAddress 类

  public class EmailAddress implements Serializable {
    private static final long serialVersionUID = -6999956021169014445L;
    private static final String AT_DELIMTER = "@";
    private String emailAddress;

    public EmailAddress() {}

    public EmailAddress(String emailAddress) {
        setEmailAddress(emailAddress);
    }

    public String getEmailAddress() {
        return emailAddress;
    }

    public void setEmailAddress(String emailAddress) throws IllegalArgumentException {
        if (StringUtils.isEmpty(emailAddress))
            throw new IllegalArgumentException("Parameter emailAddress cannot be blank.");
        else if (!EmailValidator.getInstance(true).isValid(emailAddress))
            throw new IllegalArgumentException("The email address " +  emailAddress + " is not valid.");

        this.emailAddress = emailAddress;
    }

    public String getUser() {
        return this.emailAddress.substring(0, this.emailAddress.indexOf(AT_DELIMTER));
    }

    public String getDomain() {
        return this.emailAddress.substring(this.emailAddress.indexOf(AT_DELIMTER) + 1);
    }

    public String toString() {
        return emailAddress;
    }

    @Override
    public int hashCode() {
        final int prime = 31;
        int result = 1;
        result = prime * result
                + ((emailAddress == null) ? 0 : emailAddress.hashCode());
        return result;
    }

    @Override
    public boolean equals(Object obj) {
        if (this == obj)
            return true;
        if (obj == null)
            return false;
        if (getClass() != obj.getClass())
            return false;
        EmailAddress other = (EmailAddress) obj;
        if (emailAddress == null) {
            if (other.emailAddress != null)
                return false;
        } else if (!emailAddress.equals(other.emailAddress))
            return false;
        return true;
    }
}

谢谢!

Answer 1

将您的查询更改为:

List results = this.sessionFactory.getCurrentSession()
    .createQuery("select au from AppUser au where au.email = :email")
    .setParameter("email", email)
    .list();

您需要实现 Hibernate custom UserType在加载时将数据库字符串转换为 EmailAddress。

使用 ImmutableType来自 hibernate-types 项目作为自定义 Hibernate Type 的基类，因为这样实现它要容易得多。