hibernate用户登录could not execute query 大家来找找错吧！

Question

HttpSession session = request.getSession();
Session s = HibernateUtil.currentSession(); //得到连接
LoginForm loginForm = (LoginForm) form; String sort = loginForm.getSort();
String username = loginForm.getUsername();
String password = loginForm.getPassword();
int loginSort = Integer.parseInt(sort); String[] userlist = new String[2];
userlist[0] = username;
userlist[1] = password; try {
HibernateUtil.beginTransaction();//开始事物
String str = new String();
switch (loginSort) {
case 1:
str = " from Student as stu where stu.name=:stuName and stu.password=:stuPassword";

Query query = s.createQuery(str);
//System.out.println(username + "  " + password);
query.setString("stuName", username);
query.setString("stuPassword", password);
System.out.println(str);
if (query.list().size() > 0) {
session.setAttribute("stuid", ((Student) query.list().get(0)).getId());
s.close();
return mapping.findForward("studentLoginsuccess");
} else
break;

Answer 1

<?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<!-- 
    Mapping file autogenerated by MyEclipse - Hibernate Tools
-->
<hibernate-mapping>
    <class name="com.stuman.domain.Student" table="student" lazy="false">
        <id name="id" type="string">
            <column name="id" length="32" />
            <generator class="assigned" />
        </id>
        <property name="name" type="string">
            <column name="name" length="32" />
        </property>
        <property name="password" type="string">
            <column name="password" length="32" />
        </property>
        <property name="jiguan" type="string">
            <column name="jiguan" length="32" />
        </property>
        <property name="department" type="string">
            <column name="department" length="32" />
        </property>
        <property name="sex" type="string">
            <column name="sex" length="32" />
        </property>
        <property name="" type="integer">
            <column name="" />
        </property>
        <property name="tel" type="string">
            <column name="tel" length="32" />
        </property>
        <property name="phone" type="string">
            <column name="phone" length="32" />
        </property>
        <property name="email" type="string">
            <column name="email" length="32" />
        </property>
        <set name="enrols" inverse="true">
            <key>
                <column name="stu_id" length="32" />
            </key>
            <one-to-many class="com.stuman.domain.Enrol" />
        </set>
    </class>
</hibernate-mapping>

Answer 2

请仔细检查student表中的phone列有没有错误

Answer 3

你可以把语句:
   str = " from Student as stu where stu.name=:stuName and stu.password=:stuPassword";
换成:
  str = " from Student as stu where stu.name="+username+" and stu.password="+password;
另外:
   你Loginform   loginform=(Loginform)form;
     loginform.getSort();
     是做什么的;

Answer 4

ERROR org.hibernate.util.JDBCExceptionReporter  - Unknown column 'student0_.phone' in 'field list' 

Answer 5

查查数据库里的字段，是不是phone这个字段写错了