HQL 语句 public User Login(String username) { String hql = "from User user where user.username = ? "; User user = (User) this.getHibernateTemplate().find(hql, username).get(0); return user; }action 进行判断 public String execute() throws Exception { User u = (User) userManager.Login(username); Map session = ActionContext.getContext().getSession();
return getHibernateTemplate().execute(new HibernateCallback<List<E>>() {
public List<E> doInHibernate(Session session) throws HibernateException {
Query query = session.createQuery("FROM User u WHERE u.name=:name");
query.setString("name", name);
return query.list();
}
});
希望对你有帮助。我刚工作时候写的。
<!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="connection.url">jdbc:sqlserver://localhost:1433;databaseName=hibernate</property>
<property name="connection.username">sa</property>
<property name="connection.password">svse</property>
<property name="connection.driver_class">com.microsoft.sqlserver.jdbc.SQLServerDriver</property>
<property name="dialect">org.hibernate.dialect.MySQLDialect</property>
<property name="show_sql">true</property>
<mapping resource="Peson.hbm.xml"/>
</session-factory>
</hibernate-configuration>
public User Login(String username) {
String hql = "from User user where user.username = ? ";
User user = (User) this.getHibernateTemplate().find(hql, username).get(0);
return user;
}action 进行判断
public String execute() throws Exception {
User u = (User) userManager.Login(username);
Map session = ActionContext.getContext().getSession();
if(u != null && u.getPassword().equals(password))
{
this.user = u;
ActionContext.getContext().getSession().put("login", u);
return SUCCESS;
}
this.setManager("用户名或密码错误!!");
return INPUT;
}
}