【新手求救】hql语句查询问题

buzhidao1123 2014-10-20 12:44:19

本人禽兽学习hql遇到问题请求支援

public static void main(String args[]){

		Session session = HibernateUtil.getSessionFactory().openSession();

		Query query  =  session.createQuery("from House");

		List<House> list = query.list();

		Iterator<House> it = list.iterator();

		System.out.println("我运行了");

		while(it.hasNext()){

			House house = it.next();

			System.out.println(house.getDescription()+" "+house.getTitle());

		}

		System.out.println("我运行即将结束");

		session.close();

	}

}

我的House里有数据的但是运行结果如下
Hibernate:
select
house0_.id as id3_,
house0_.user_id as user2_3_,
house0_.type_id as type3_3_,
house0_.street_id as street4_3_,
house0_.title as title3_,
house0_.description as descript6_3_,
house0_.price as price3_,
house0_.pubdate as pubdate3_,
house0_.adddate as adddate3_,
house0_.floorage as floorage3_,
house0_.contact as contact3_
from
house house0_
我运行了
我运行即将结束
-------------------------------------------------------------
为什么迭代不出东西来呢？

...全文

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littlebrain4solving 2014-10-20

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引用 10 楼 buzhidao1123 的回复:

[quote=引用 9 楼 littlebrain4solving 的回复:] 循环进入了吗?List的size大小查看一下,你打断点单步调试就能看的清清楚楚! 用Debug启动

发现了是没有数据没有权限给了权限就好了另外问一下运行后出现这个错误 No row with the given identifier exists: [com.pb.entity.HouseUser#5] 我检查不出来了

<?xml version="1.0" encoding="utf-8"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<hibernate-mapping>
	<class name="com.pb.entity.House" table="house" lazy="false">
		<id name="id" type="int">
			<column name="id" />
			<generator class="assigned">
			</generator>
		</id>
		<many-to-one name="houseUser" class="com.pb.entity.HouseUser"
			cascade="none" fetch="join">
			<column name="houseuser_id" />
		</many-to-one>
		<many-to-one name="houseType" class="com.pb.entity.HouseType"
			cascade="none" fetch="join">
			<column name="type_id" />
		</many-to-one>
		<many-to-one name="street" class="com.pb.entity.Street"
			cascade="none" fetch="join">
			<column name="street_id" />
		</many-to-one>
		<property name="title" type="string">
			<column name="title" length="50" not-null="false" />
		</property>
		<property name="description" type="string" lazy="false">
			<column name="description" />
		</property>
		<property name="price" type="double">
			<column name="price" precision="10" scale="4" not-null="false" />
		</property>
		<property name="pubdate" type="date">
			<column name="pubdate" length="23" />
		</property>
		<property name="adddate" type="date">
		    <column name="adddate" length="23" />
		</property>
		<property name="floorage" type="double">
			<column name="floorage" length="10" />
		</property>
		<property name="contact" type="string">
			<column name="contact" length="10" />
		</property>

	</class>
	</hibernate-mapping>

、、、、、、、、、、、、、、、、、、、、、、、、、、、、、、、、、、、、这是house 、、、、、、、、、、、、、、、、、、、、、、、、、、、、、、、、、、、、、

<?xml version="1.0" encoding="utf-8"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<hibernate-mapping>
	<class name="com.pb.entity.HouseUser" table="HOUSEUSER" lazy="false">
		<id name="id" type="int">
			<column name="id" />
			<generator class="sequence" >
            <param name="sequence">house_user</param>
            </generator>
		</id>
	<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="telephone" type="string">
			<column name="telephone"/>
	</property>
	<property name="isadmin" type="string">
			<column name="isadmin"/>
	</property>
	<property name="realname" type="string">
			<column name="realname"/>
		
	</property>
	</class>
</hibernate-mapping>

这是HouseUser[/quote] 参考资料 http://zhidao.baidu.com/link?url=jIW0Eg6Reu4r9PC-F_EgVEBkBGqZYKfpOTnSI1qDYdr8ixaCq4RojCCX0FRSHq92IROYI4HyXCi6csiqDgWJ-K http://blog.csdn.net/eyejava/article/details/1896492

shixitong 2014-10-20

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1、插入的数据确认commit了吧（在数据库中存在）？ 2、获取session改为这样试试

Session session = HibernateUtil.getSessionFactory().getSession();

buzhidao1123 2014-10-20

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引用 9 楼 littlebrain4solving 的回复:

循环进入了吗?List的size大小查看一下,你打断点单步调试就能看的清清楚楚! 用Debug启动

发现了是没有数据没有权限给了权限就好了另外问一下运行后出现这个错误 No row with the given identifier exists: [com.pb.entity.HouseUser#5] 我检查不出来了

<?xml version="1.0" encoding="utf-8"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<hibernate-mapping>
	<class name="com.pb.entity.House" table="house" lazy="false">
		<id name="id" type="int">
			<column name="id" />
			<generator class="assigned">
			</generator>
		</id>
		<many-to-one name="houseUser" class="com.pb.entity.HouseUser"
			cascade="none" fetch="join">
			<column name="houseuser_id" />
		</many-to-one>
		<many-to-one name="houseType" class="com.pb.entity.HouseType"
			cascade="none" fetch="join">
			<column name="type_id" />
		</many-to-one>
		<many-to-one name="street" class="com.pb.entity.Street"
			cascade="none" fetch="join">
			<column name="street_id" />
		</many-to-one>
		<property name="title" type="string">
			<column name="title" length="50" not-null="false" />
		</property>
		<property name="description" type="string" lazy="false">
			<column name="description" />
		</property>
		<property name="price" type="double">
			<column name="price" precision="10" scale="4" not-null="false" />
		</property>
		<property name="pubdate" type="date">
			<column name="pubdate" length="23" />
		</property>
		<property name="adddate" type="date">
		    <column name="adddate" length="23" />
		</property>
		<property name="floorage" type="double">
			<column name="floorage" length="10" />
		</property>
		<property name="contact" type="string">
			<column name="contact" length="10" />
		</property>

	</class>
	</hibernate-mapping>

<?xml version="1.0" encoding="utf-8"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<hibernate-mapping>
	<class name="com.pb.entity.HouseUser" table="HOUSEUSER" lazy="false">
		<id name="id" type="int">
			<column name="id" />
			<generator class="sequence" >
            <param name="sequence">house_user</param>
            </generator>
		</id>
	<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="telephone" type="string">
			<column name="telephone"/>
	</property>
	<property name="isadmin" type="string">
			<column name="isadmin"/>
	</property>
	<property name="realname" type="string">
			<column name="realname"/>
		
	</property>
	</class>
</hibernate-mapping>

这是HouseUser

littlebrain4solving 2014-10-20

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循环进入了吗?List的size大小查看一下,你打断点单步调试就能看的清清楚楚! 用Debug启动

buzhidao1123 2014-10-20

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引用 4 楼 shixitong 的回复:

select * from House在数据库里执行下看看，如果没有数据，加几条数据就行了

有数据的为了确定还新添加了2条呢 insert into house values(1,'北京好房！不买后悔','传统的两室一厅，光线好，格局好',1000,null,null,132,'好地方',1,2,3); insert into house values(2,'通州房屋,仅剩3款','超大面积享受，四室三厅，开窗见河，风景优美',200000,null,null,122,'好风光',1,2,3); insert into house values(3,'王府井大街新楼开盘','全新体验，全新享受，高档小区',40000,null,null,111,'行不行',1,2,2); insert into house values(4,'西单新房','距地铁近，好氛围，好景色，便宜',300000,null,null,311,'很好的',2,3,5);