我在使用 Struts2 和 Hibernate 开发 Web 应用程序时遇到了以下 Exception.am 问题。有人可以帮助我解决此问题吗?
<小时/>org.hibernate.MappingException: Unknown entity: Pojo.USER [Packagename.entityClassname]
Hibernate.cfg.xml 文件内容:-
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE hibernate-configuration PUBLIC
"-//Hibernate/Hibernate Configuration DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-configuration-3.0.dtd">
<hibernate-configuration>
<session-factory>
<property name="dialect">org.hibernate.dialect.MySQLDialect</property>
<property name="connection.driver_class">com.mysql.jdbc.Driver</property>
<property name="connection.url">jdbc:mysql://localhost:3306/databasename
</property>
<property name="connection.username">root</property>
<property name="connection.password">admin</property>
<property name="cache.provider_class">
org.hibernate.cache.HashtableCacheProvider</property>
<property name="transaction.factory_class">
org.hibernate.transaction.JDBCTransactionFactory</property>
<property name="current_session_context_class">thread</property>
<property name="hibernate.show_sql">false</property>
</session-factory>
</hibernate-configuration>
<小时/>
用户的实体类内容:-
package Pojo;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.Id;
import javax.persistence.Table;
@Entity
@Table(name="user")
public class USER {
@Id
@GeneratedValue
@Column(name="USER_ID")
private int id;
@Column(name="USER_NAME")
private String name;
@Column(name="USER_PASSWORD")
private String password;
@Column(name="USER_GENDER")
private String gender;
@Column(name="USER_COUNTRY")
private String country;
@Column(name="USER_ABOUT_YOU")
private String aboutyou;
@Column(name="USER_MAILING_LIST")
private boolean mailinglist;
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getPassword() {
return password;
}
public void setPassword(String password) {
this.password = password;
}
public String getGender() {
return gender;
}
public void setGender(String gender) {
this.gender = gender;
}
public String getCountry() {
return country;
}
public void setCountry(String country) {
this.country = country;
}
public String getAboutyou() {
return aboutyou;
}
public void setAboutyou(String aboutyou) {
this.aboutyou = aboutyou;
}
public boolean isMailinglist() {
return mailinglist;
}
public void setMailinglist(boolean mailinglist) {
this.mailinglist = mailinglist;
}
}
最佳答案
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE hibernate-configuration PUBLIC
"-//Hibernate/Hibernate Configuration DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-configuration-3.0.dtd">
<hibernate-configuration>
<session-factory>
<property name="dialect">org.hibernate.dialect.MySQLDialect</property>
<property name="connection.driver_class">com.mysql.jdbc.Driver</property>
<property name="connection.url">jdbc:mysql://localhost:3306/databasename
</property>
<property name="connection.username">root</property>
<property name="connection.password">admin</property>
<property name="cache.provider_class">
org.hibernate.cache.HashtableCacheProvider</property>
<property name="transaction.factory_class">
org.hibernate.transaction.JDBCTransactionFactory</property>
<property name="current_session_context_class">thread</property>
<property name="hibernate.show_sql">false</property>
<mapping class="Pojo.User">
</session-factory>
</hibernate-configuration>
使用映射类,这样hibernate就会知道你的实体类--> <mapping class="Pojo.User">
关于java - 如何解决 "org.hibernate.MappingException: Unknown entity"错误?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/29794318/