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List的removeAll方法错误

发布时间: 2012-04-01 17:23:46 作者: rapoo

List的removeAll方法异常?
现在正在实现的一个模块,是用Hibernate做的,一个用户对应多个频道,用的是一对多的关联,配置方件如下:
User表:

Java code
<?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 lazy="false" name="com.autonavi.monitor.model.User" table="user">  <id name="userId" type="integer" unsaved-value="0">   <column name="userId"/>   <generator class="native"/>  </id>  <property  name="name" type="string">   <column name="name" />  </property>  <property  name="userName" type="string">   <column name="userName" not-null="true"/>  </property>  <property name="password" type="string">   <column name="password" not-null="true"/>  </property>  <property name="actor" type="int">   <column name="actor"/>  </property>  <property name="authority" type="string">   <column name="authority"/>  </property>  <property  name="email" type="string">   <column name="email"/>  </property>  <property name="phone" type="string">   <column name="phone"/>  </property> <property name="phone1" type="string">   <column name="phone1"/>  </property>  <property name="phone2" type="string">   <column name="phone2"/>  </property>    <set name="channels" cascade="all" lazy="false">      <key column="userId" />      <one-to-many class="com.autonavi.monitor.model.Channel"/>  </set>   </class></hibernate-mapping>


频道表:
Java code
<?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 lazy="false" name="com.autonavi.monitor.model.Channel" table="channel">  <id name="channelId" type="integer" unsaved-value="0">   <column name="channelId"/>   <generator class="native"/>  </id>  <property  name="channel" type="string">   <column name="channel" />  </property>  <property  name="channelName" type="string">   <column name="channel_name"/>  </property>  <property  name="delayTime" type="integer">   <column name="delay_time" not-null="true"/>  </property> </class></hibernate-mapping>

在Action里面做相应的查询,由于取消了懒加载,在查出的User里面会取到对应的频道列表
Java code
List<Channel> userChannelList = new ArrayList<Channel>();            Channel channel = null;            Set<Channel> channelSet = modifyUser.getChannels();            Iterator<Channel> iterator = channelSet.iterator();            while(iterator.hasNext()) {                channel = (Channel)iterator.next();                userChannelList.add(channel);            } 



用Hibernate的HQL功能取到所有的频道:
Java code
@SuppressWarnings("unchecked")    public List<Channel> getChannelList() {        String sql = "";        sql = "from Channel";        List<Channel> list = this.getHibernateTemplate().find(sql);        return list;    }


现在我要得到剩下的频道,所以想着是用得到的所有的频道List调用removeAll方法去掉当前用户占用的频道List
Java code
channelList.removeAll(userChannelList);


本来挺简单的,可执行后发现并没有达到预期的效果,不能从所有的频道中把当前用户占用的频道remove掉,找了很久才发现所有频道List中包含的当前用户占用的频道的hashcode和当前用户占用的频道List中的频道的hashcode值不一样,其他值都相同,猜测可能是因为这所以不能执行remove操作,我想知道为什么会出现这种情况?该怎么实现我上面所说的需求?




[解决办法]
hashcode算法加入了数组的初始长度,也就是说初始长度不同的数组(ArrayList)中相同对象的hashcode有可能不同
[解决办法]
channelList只能remove自己的对象,你那个list是别的对象,它怎么删别人?
你只有通过嵌套for,根据判断channelList.get(i).channelId==userChannelList.get(j).channelId
如果相同,则channelList.remove(i),这样一个一个删除。

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