本文介绍了`packagesToScan`不适用于@Entity和@NamedQuery的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
这是我的User.java类
package com.sudosmith.cenmesser.model;
import java.io.Serializable;
import java.util.ArrayList;
import java.util.List;
import javax.persistence.Column;
import javax.persistence.Embeddable;
import javax.persistence.Entity;
import javax.persistence.EnumType;
import javax.persistence.Enumerated;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.JoinTable;
import javax.persistence.ManyToMany;
import javax.persistence.NamedQuery;
import javax.persistence.Table;
$ b $ **
*
* @author Vishal Joshi
*
* /
@ Entity
@Table（name =User）
@NamedQuery（name =findUserByName，query =SELECT user FROM User user WHERE user.userName LIKE：queryString）
@ org.hibernate.annotations。实体（dynamicInsert = true，dynamicUpdate = true）
public class User实现Serializable {
$ b $ / **
*
* /
private static final long serialVersionUID = 7257051297399186919L;
@Id
@GeneratedValue（strategy = GenerationType.AUTO）
private int id;
@Column（name =username）
private String userName;
@Column（name =password）
private String password;
$ b @ManyToMany
@JoinTable（name =user_role，joinColumns = @JoinColumn（name =userid），inverseJoinColumns = @JoinColumn（name =roleid））
私人列表< Role>角色;
@Enumerated（EnumType.STRING）
@Column（name =status）
私有UserStatus状态;
public User（）{
}
public User（int id，String userName，String password，List< Role> roles，
UserStatus状态）{
super（）;
this.id = id;
this.userName = userName;
this.password =密码;
this.roles =角色;
this.status = status;
$ b $ **
* @return the id
* /
public int getId（）{
return id;
$ b $ ** b $ b * @param id
*要设置的ID
* /
public void setId（int id） {
this.id = id;
$ b / **
* @return userName
* /
public String getUserName（）{
return userName;
}
$ b / **
* @param userName
*用户名设置
* /
public void setUserName（String userName） {
this.userName = userName;
$ b / **
* @返回密码
* /
public String getPassword（）{
return password;
}
$ b / **
* @param密码
*密码设置
* /
public void setPassword（String password） {
this.password = password;
}
/ **
* @返回角色
* /
public List< Role> getRoles（）{
if（roles！= null）{
返回角色;
} else {
返回新的ArrayList< Role>（）;
}
}
/ **
* @param角色
*设置角色
* /
public void setRoles（列出<角色>角色）{
this.roles =角色;
$ b / **
* @返回状态
* /
public UserStatus getStatus（）{
return status;
}
$ b $ **
* @param status
*要设置的状态
* /
public void setStatus（UserStatus status） {
this.status = status;
}
}
在我的 spring-servlet.xml 我已经使用 packagesToScan 属性来映射所有具有 @Entity
< bean id =sessionFactoryclass =org.springframework.orm.hibernate3。 annotation.AnnotationSessionFactoryBean>
< property name =dataSourceref =dataSource/>
< property name =hibernateProperties>
<道具>
< prop key =hibernate.dialect> $ {jdbc.dialect}< / prop>
< prop key =hibernate.show_sql> true< / prop>
< /道具>
< / property> < property name =packagesToScan>
< list>
<值> com.sudosmith.cenmesser.model。**< /值>
< / list>
< / property>
< / bean>
但是在添加此属性后，它不起作用。它不映射 @Entity 和 @NamedQuery 。我遇到了错误，如 mappingException：unknown @NamedQuery 。
$ b 编辑：Hibernate依赖关系
<依赖关系>
< groupId> org.hibernate< / groupId>
< artifactId> hibernate-entitymanager< / artifactId>
< version> 3.6.7.Final< / version>
< /依赖关系>
< dependency>
< groupId> org.hibernate< / groupId>
< artifactId> hibernate-validator< / artifactId>
< version> 4.3.0.Final< / version>
< /依赖关系>
< dependency>
< groupId> org.hibernate< / groupId>
< artifactId> hibernate-annotations< / artifactId>
< version> 3.3.1.GA< / version>
< /依赖关系>
< dependency>
< groupId> org.hibernate< / groupId>
< artifactId> hibernate-core< / artifactId>
< version> 4.2.11< / version>
< /依赖关系>
解决方案
使用<值> com.sudosmith.cenmesser.model< /值>
This is my User.java class
package com.sudosmith.cenmesser.model;
import java.io.Serializable;
import java.util.ArrayList;
import java.util.List;
import javax.persistence.Column;
import javax.persistence.Embeddable;
import javax.persistence.Entity;
import javax.persistence.EnumType;
import javax.persistence.Enumerated;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.JoinTable;
import javax.persistence.ManyToMany;
import javax.persistence.NamedQuery;
import javax.persistence.Table;
/**
*
* @author Vishal Joshi
*
*/
@Entity
@Table(name = "User")
@NamedQuery(name = "findUserByName", query = "SELECT user FROM User user WHERE user.userName LIKE :queryString")
@org.hibernate.annotations.Entity(dynamicInsert = true, dynamicUpdate = true)
public class User implements Serializable {
/**
*
*/
private static final long serialVersionUID = 7257051297399186919L;
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private int id;
@Column(name = "username")
private String userName;
@Column(name = "password")
private String password;
@ManyToMany
@JoinTable(name = "user_role", joinColumns = @JoinColumn(name = "userid"), inverseJoinColumns = @JoinColumn(name = "roleid"))
private List<Role> roles;
@Enumerated(EnumType.STRING)
@Column(name = "status")
private UserStatus status;
public User() {
}
public User(int id, String userName, String password, List<Role> roles,
UserStatus status) {
super();
this.id = id;
this.userName = userName;
this.password = password;
this.roles = roles;
this.status = status;
}
/**
* @return the id
*/
public int getId() {
return id;
}
/**
* @param id
* the id to set
*/
public void setId(int id) {
this.id = id;
}
/**
* @return the userName
*/
public String getUserName() {
return userName;
}
/**
* @param userName
* the userName to set
*/
public void setUserName(String userName) {
this.userName = userName;
}
/**
* @return the password
*/
public String getPassword() {
return password;
}
/**
* @param password
* the password to set
*/
public void setPassword(String password) {
this.password = password;
}
/**
* @return the roles
*/
public List<Role> getRoles() {
if (roles != null) {
return roles;
} else {
return new ArrayList<Role>();
}
}
/**
* @param roles
* the roles to set
*/
public void setRoles(List<Role> roles) {
this.roles = roles;
}
/**
* @return the status
*/
public UserStatus getStatus() {
return status;
}
/**
* @param status
* the status to set
*/
public void setStatus(UserStatus status) {
this.status = status;
}
}
in my spring-servlet.xml I have used the packagesToScan property to map all the model classes which has @Entity annotation.
<bean id="sessionFactory" class="org.springframework.orm.hibernate3.annotation.AnnotationSessionFactoryBean">
<property name="dataSource" ref="dataSource" />
<property name="hibernateProperties">
<props>
<prop key="hibernate.dialect">${jdbc.dialect}</prop>
<prop key="hibernate.show_sql">true</prop>
</props>
</property> <property name="packagesToScan">
<list>
<value>com.sudosmith.cenmesser.model.**</value>
</list>
</property>
</bean>
But after adding this property its not working. Its not mapping @Entity and @NamedQuery. I am getting error like mappingException: unknown @NamedQuery.
EDIT: Hibernate Dependencies
<dependency>
<groupId>org.hibernate</groupId>
<artifactId>hibernate-entitymanager</artifactId>
<version>3.6.7.Final</version>
</dependency>
<dependency>
<groupId>org.hibernate</groupId>
<artifactId>hibernate-validator</artifactId>
<version>4.3.0.Final</version>
</dependency>
<dependency>
<groupId>org.hibernate</groupId>
<artifactId>hibernate-annotations</artifactId>
<version>3.3.1.GA</version>
</dependency>
<dependency>
<groupId>org.hibernate</groupId>
<artifactId>hibernate-core</artifactId>
<version>4.2.11</version>
</dependency>
解决方案
Try it with <value>com.sudosmith.cenmesser.model</value>
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