我正在尝试从"用户"表中获取所有用户的列表,我收到以下错误:
org.hibernate.hql.internal.ast.QuerySyntaxException: users is not mapped [from users]
org.hibernate.hql.internal.ast.util.SessionFactoryHelper.requireClassPersister(SessionFactoryHelper.java:180)
org.hibernate.hql.internal.ast.tree.FromElementFactory.addFromElement(FromElementFactory.java:110)
org.hibernate.hql.internal.ast.tree.FromClause.addFromElement(FromClause.java:93)
Run Code Online (Sandbox Code Playgroud)
这是我写的添加/获取用户的代码:
public List<User> getUsers() {
Session session = HibernateUtil.getSessionFactory().getCurrentSession();
session.beginTransaction();
List<User> result = (List<User>) session.createQuery("from users").list();
session.getTransaction().commit();
return result;
}
public void addUser(User user) {
Session session = HibernateUtil.getSessionFactory().getCurrentSession();
session.beginTransaction();
session.save(user);
session.getTransaction().commit();
}
public void addUser(List<User> users) {
Session session = HibernateUtil.getSessionFactory().getCurrentSession();
session.beginTransaction();
for (User user : users) {
session.save(user);
}
session.getTransaction().commit();
}
Run Code Online (Sandbox Code Playgroud)
添加用户是有效的,但是当我使用getUsers函数时,我收到了这些错误.
这是我的hibernate配置文件:
<hibernate-configuration>
<session-factory>
<property name="connection.url">jdbc:mysql://localhost:3306/test</property>
<property name="connection.username">root</property>
<property name="connection.password">root</property>
<property name="connection.driver_class">com.mysql.jdbc.Driver</property>
<property name="hibernate.default_schema">test</property>
<property name="dialect">org.hibernate.dialect.MySQL5Dialect</property>
<property name="show_sql">true</property>
<property name="format_sql">true</property>
<property name="hbm2ddl.auto">create-drop</property>
<!-- JDBC connection pool (use the built-in) -->
<property name="connection.pool_size">1</property>
<property name="current_session_context_class">thread</property>
<!-- Mapping files will go here.... -->
<mapping class="model.Company" />
<mapping class="model.Conference" />
<mapping class="model.ConferencesParticipants" />
<mapping class="model.ConferenceParticipantStatus" />
<mapping class="model.ConferencesUsers" />
<mapping class="model.Location" />
<mapping class="model.User" />
</session-factory>
Run Code Online (Sandbox Code Playgroud)
这是我的User类:
@Entity
@Table( name = "Users" )
public class User implements Serializable{
private long userID;
private int pasportID;
private Company company;
private String name;
private String email;
private String phone1;
private String phone2;
private String password; //may be null/empty , will be kept hashed
private boolean isAdmin;
private Date lastLogin;
User() {} //not public on purpose!
public User(int countryID, Company company, String name, String email,
String phone1, String phone2, String password, boolean isAdmin) {
this.pasportID = countryID;
this.company = company;
this.name = name;
this.email = email;
this.phone1 = phone1;
this.phone2 = phone2;
this.password = password;
this.isAdmin = isAdmin;
}
@Id
@GeneratedValue(generator="increment")
@GenericGenerator(name="increment", strategy = "increment")
public long getUserID() {
return userID;
}
public void setUserID(long userID) {
this.userID = userID;
}
...
}
Run Code Online (Sandbox Code Playgroud)
知道为什么我会收到这个错误吗?
Ken*_*han 279
在HQL中,您应该使用映射的java类名和属性名@Entity而不是实际的表名和列名,因此HQL应该是:
List<User> result = (List<User>) session.createQuery("from User").list();
Run Code Online (Sandbox Code Playgroud)
Kum*_*mal 27
例如:您的bean类名称是UserDetails
Query query = entityManager. createQuery("Select UserName from **UserDetails** ");
Run Code Online (Sandbox Code Playgroud)
您没有在Db上提供您的表名.你给bean的类名.
Iwa*_*ria 12
只是为了分享我的发现.即使查询的目标是正确的类名,我仍然会遇到相同的错误.后来我意识到我从错误的包中导入了Entity类.
更改导入行后,问题解决了:
import org.hibernate.annotations.Entity;
Run Code Online (Sandbox Code Playgroud)
至
import javax.persistence.Entity;
Run Code Online (Sandbox Code Playgroud)
添加了@TABLE(name = "TABLE_NAME")注释并修复了.检查您的注释和hibernate.cfg.xml文件.这是有效的示例实体文件:
import javax.persistence.*;
@Entity
@Table(name = "VENDOR")
public class Vendor {
//~ --- [INSTANCE FIELDS] ------------------------------------------------------------------------------------------
private int id;
private String name;
//~ --- [METHODS] --------------------------------------------------------------------------------------------------
@Override
public boolean equals(final Object o) {
if (this == o) {
return true;
}
if (o == null || getClass() != o.getClass()) {
return false;
}
final Vendor vendor = (Vendor) o;
if (id != vendor.id) {
return false;
}
if (name != null ? !name.equals(vendor.name) : vendor.name != null) {
return false;
}
return true;
}
//~ ----------------------------------------------------------------------------------------------------------------
@Column(name = "ID")
@GeneratedValue(strategy = GenerationType.AUTO)
@Id
public int getId() {
return id;
}
@Basic
@Column(name = "NAME")
public String getName() {
return name;
}
public void setId(final int id) {
this.id = id;
}
public void setName(final String name) {
this.name = name;
}
@Override
public int hashCode() {
int result = id;
result = 31 * result + (name != null ? name.hashCode() : 0);
return result;
}
}
Run Code Online (Sandbox Code Playgroud)
org.hibernate.hql.internal.ast.QuerySyntaxException: users is not mapped [from users]
这表明hibernate不知道该User实体是"用户".
@javax.persistence.Entity
@javax.persistence.Table(name = "Users")
public class User {
Run Code Online (Sandbox Code Playgroud)
该@Table注释设置表的名称是"用户",但实体名称仍然指在HQL"用户".
要更改两者,您应该设置实体的名称:
// this sets the name of the table and the name of the entity
@javax.persistence.Entity(name = "Users")
public class User implements Serializable{
Run Code Online (Sandbox Code Playgroud)
有关详细信息,请参阅此处的答案:Hibernate表未映射错误
另请检查您是否使用以下命令添加了带注释的类:
new Configuration().configure("configuration file path").addAnnotatedClass(User.class)
使用 Hibernate 在数据库中添加新表时,这总是浪费我的时间。
小智 5
还要确保在hibernate bean配置中设置了以下属性:
<property name="packagesToScan" value="yourpackage" />
Run Code Online (Sandbox Code Playgroud)
这告诉spring和hibernate在哪里找到注释为实体的域类.
小智 5
一些基于Linux的MySQL安装需要区分大小写.解决方法是申请nativeQuery.
@Query(value = 'select ID, CLUMN2, CLUMN3 FROM VENDOR c where c.ID = :ID', nativeQuery = true)
Run Code Online (Sandbox Code Playgroud)
| 归档时间: |
|
| 查看次数: |
216736 次 |
| 最近记录: |