如何使用Hibernate
我想从表中得到一个单一的价值,但在那里如何使用Hibernate
有错误报道让我考虑一个简单的SQL语法来获得从表中的用户ID。
选择从tbl_name其中电子邮件= '[email protected]' 身份证;
现在我想返回使用用户ID的方法冬眠
这里是我迄今为止
public int getIdByEmail(String email) { session = sessionFact.openSession();
Query query = session.createQuery("SELECT u.user_id FROM tbl_user u WHERE u.email=:emailParam");
query.setParameter("emailParam", email);
return (int) query.uniqueResult();
}
试图User.java
@Entity @Table(name = "tbl_user", catalog = "lifestyle", schema = "")
@NamedQueries({
@NamedQuery(name = "User.findAll", query = "SELECT u FROM User u")
, @NamedQuery(name = "User.findByUserId", query = "SELECT u FROM User u WHERE u.userId = :userId")
, @NamedQuery(name = "User.findByFullName", query = "SELECT u FROM User u WHERE u.fullName = :fullName")
, @NamedQuery(name = "User.findByAddress", query = "SELECT u FROM User u WHERE u.address = :address")
, @NamedQuery(name = "User.findByContact", query = "SELECT u FROM User u WHERE u.contact = :contact")
, @NamedQuery(name = "User.findByGender", query = "SELECT u FROM User u WHERE u.gender = :gender")
, @NamedQuery(name = "User.findByDob", query = "SELECT u FROM User u WHERE u.dob = :dob")
, @NamedQuery(name = "User.findByEmail", query = "SELECT u FROM User u WHERE u.email = :email")
, @NamedQuery(name = "User.findByPassword", query = "SELECT u FROM User u WHERE u.password = :password")
, @NamedQuery(name = "User.findByActive", query = "SELECT u FROM User u WHERE u.active = :active")
, @NamedQuery(name = "User.findByCreatedDate", query = "SELECT u FROM User u WHERE u.createdDate = :createdDate")})
public class User implements Serializable {
private static final long serialVersionUID = 1L;
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "user_id")
private Integer userId;
@NotNull
@Size(min = 1, max = 256)
@Column(name = "full_name")
private String fullName;
@Size(max = 256)
@Column(name = "address")
private String address;
@Size(max = 30)
@Column(name = "contact")
private String contact;
@Size(max = 10)
@Column(name = "gender")
private String gender;
@Column(name = "dob")
@Temporal(TemporalType.DATE)
private Date dob;
// @Pattern(regexp="[a-z0-9!#$%&'*+/=?^_`{|}~-]+(?:\\.[a-z0-9!#$%&'*+/=?^_`{|}~-]+)*@(?:[a-z0-9](?:[a-z0-9-]*[a-z0-9])?\\.)+[a-z0-9](?:[a-z0-9-]*[a-z0-9])?", message="Invalid email")//if the field contains email address consider using this annotation to enforce field validation
@NotNull
@Size(min = 1, max = 256)
@Column(name = "email")
private String email;
@NotNull
@Size(min = 1, max = 256)
@Column(name = "password")
private String password;
@NotNull
@Column(name = "active", insertable = false)
private short active;
@Column(name = "created_date", insertable = false)
@Temporal(TemporalType.TIMESTAMP)
private Date createdDate;
错误是
回答:
我们需要在查询中使用实体名称和属性,而不是表名和列名。因此,以下查询:
查询的查询= session.createQuery( “选择u.user_id FROM tbl_userù WHERE u.email =:emailParam”);
应该是
查询的查询= session.createQuery( “选择u.userId FROM用户U WHERE u.email =:emailParam”);
以上是 如何使用Hibernate 的全部内容, 来源链接: utcz.com/qa/266260.html