hibernate关联查询出现java.lang.NullPointerException的原因，急，在线等

Question

hSession = HibernateUtilPlus.currentSession(); 打出 hSession 值看看
Query query = hSession.createQuery(hql); 如果确实是这句抛出异常，，有可能  hSession＝＝null

Answer 1

你的hibernate配置文件中的打印sql语句的状态是开吗？ 如果是，看看控制台该语句，复制并在SQLServer中执行一下，看看结果是什么如果状态没开，将其打开，便于开发调试

Answer 2

hql中的各个class属性名字写的不对也出向这个问题

Answer 3

hSession不是null，hibernate配置文件中的打印sql语句的状态是开的，但是sql语句根本就还没解析出来，应该是解析的时候就出了问题。我数据库是oracle9i

Answer 4

<?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC
    "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
    "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd" ><hibernate-mapping>
<!--
    Created by the Middlegen Hibernate plugin 2.1    http://boss.bekk.no/boss/middlegen/
    http://www.hibernate.org/
--><class
    name="com.monitorcc.system.db.table.TabUser"
    table="TAB_USER"
>
    <meta attribute="class-description" inherit="false">
       @hibernate.class
        table="TAB_USER"
    </meta>    <id
        name="userId"
        type="int"
        column="USER_ID"
        access="field"
    >
        <meta attribute="field-description">
           @hibernate.id
            generator-class="sequence"
            type="int"
            column="USER_ID"
            access="field"
           @hibernate.generator-param
        name="sequence"
        value="SEQ_USER"
        </meta>
        <meta attribute="use-in-equals">true</meta>
        <generator class="sequence">
            <param name="sequence">SEQ_USER</param>
        </generator>
    </id>    <property
        name="userName"
        type="java.lang.String"
        column="USER_NAME"
        not-null="true"
        length="10"
    >
        <meta attribute="field-description">
           @hibernate.property
            column="USER_NAME"
            length="10"
            not-null="true"
        </meta>
    </property>
    <property
        name="userLoginName"
        type="java.lang.String"
        column="USER_LOGIN_NAME"
        length="20"
    >
        <meta attribute="field-description">
           @hibernate.property
            column="USER_LOGIN_NAME"
            length="20"
        </meta>
    </property>
    <property
        name="userLoginPwd"
        type="java.lang.String"
        column="USER_LOGIN_PWD"
        length="40"
    >
        <meta attribute="field-description">
           @hibernate.property
            column="USER_LOGIN_PWD"
            length="40"
        </meta>
    </property>
    <property
        name="userPhoneNo"
        type="java.lang.String"
        column="USER_PHONE_NO"
        length="15"
    >
        <meta attribute="field-description">
           @hibernate.property
            column="USER_PHONE_NO"
            length="15"
        </meta>
    </property>
    <property
        name="userEmail"
        type="java.lang.String"
        column="USER_EMAIL"
        length="50"
    >
        <meta attribute="field-description">
           @hibernate.property
            column="USER_EMAIL"
            length="50"
        </meta>
    </property>
    <property
        name="userStatus"
        type="int"
        column="USER_STATUS"
        length="1"
    >
        <meta attribute="field-description">
           @hibernate.property
            column="USER_STATUS"
            length="1"
        </meta>
    </property>    <!-- Associations -->    <!-- bi-directional one-to-many association to TabPrivilege -->
    <set
        name="tabPrivileges"
        lazy="true"
        inverse="true"
cascade="none"
    >
        <meta attribute="field-description">
           @hibernate.set
            lazy="true"
            inverse="true"
            cascade="all"           @hibernate.collection-key
            column="USER_ID"           @hibernate.collection-one-to-many
            class="com.monitorcc.system.db.table.TabPrivilege"
        </meta>
        <key>
            <column name="USER_ID" />
        </key>
        <one-to-many
            class="com.monitorcc.system.db.table.TabPrivilege"
        />
    </set></class>
</hibernate-mapping>
这是我表的映射文件，class属性的名字应该没写错啊

Answer 5

from TabUser as user inner join user.userId
userId不是一张表吧？