Question

我有两张桌子表1 用户（ID *（PK）的，姓名，电子邮件，PWD）表2 鸣叫（tweet_id，消息，USER_ID （FK）*）

user_id是用户表id的外键。我想在hql中使用内连接来加入这两个表。

我收到此错误。

线程中的异常＆＃34; main＆＃34; org.hibernate.hql.ast.QuerySyntaxException：意外令牌：在第1行第74列[select user_name，来自com.csc.project.user的消息e内部连接推文p在p.user_id = e.id其中e.user_name =：名称] 在org.hibernate.hql.ast.QuerySyntaxException.convert（QuerySyntaxException.java:31）在org.hibernate.hql.ast.QuerySyntaxException.convert（QuerySyntaxException.java:24）在org.hibernate.hql.ast.ErrorCounter.throwQueryException（ErrorCounter.java:59）在org.hibernate.hql.ast.QueryTranslatorImpl.parse（QueryTranslatorImpl.java:258）在org.hibernate.hql.ast.QueryTranslatorImpl.doCompile（QueryTranslatorImpl.java:157）在org.hibernate.hql.ast.QueryTranslatorImpl.compile（QueryTranslatorImpl.java:111）在org.hibernate.engine.query.HQLQueryPlan。（HQLQueryPlan.java:77）在org.hibernate.engine.query.HQLQueryPlan。（HQLQueryPlan.java:56）在org.hibernate.engine.query.QueryPlanCache.getHQLQueryPlan（QueryPlanCache.java:72） at org.hibernate.impl.AbstractSessionImpl.getHQLQueryPlan（AbstractSessionImpl.java:133）在org.hibernate.impl.AbstractSessionImpl.createQuery（AbstractSessionImpl.java:112）在org.hibernate.impl.SessionImpl.createQuery（SessionImpl.java:1623）在com.csc.project.tweetManager.viewtweet（tweetManager.java:80）在com.csc.project.tweetManager.main（tweetManager.java:103）

这是我的hibernate.cfg

<!DOCTYPE hibernate-configuration PUBLIC
"-//Hibernate/Hibernate Configuration DTD 3.0//EN"
"hibernate-configuration-3.0.dtd">

<hibernate-configuration>
<session-factory>


<property name="connection.driver_class">com.mysql.jdbc.Driver</property>
<property name="connection.url">jdbc:mysql://localhost/twitter</property>
<property name="connection.user">root</property>
<property name="connection.password">root</property>



<property name="show_sql">true </property>
<property name="dialet">org.hibernate.dialect.MySQLDialect </property>
<property name="hbm2ddl.auto">update </property>

<mapping resource="registration.hbm.xml" />
<mapping resource="tweet.hbm.xml"/>


</session-factory>
</hibernate-configuration>

reigistration.hbm

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE hibernate-mapping PUBLIC 
 "-//Hibernate/Hibernate Mapping DTD//EN"
 "hibernate-mapping-3.0.dtd">
<hibernate-mapping>
   <class name="com.csc.project.user" table="t1">
      <meta attribute="class-description">
         This class contains the employee detail. 
      </meta>
      <id name="id" type="int" column="id">
         <generator class="increment"/>
      </id>



      <property name="user_name" column="user_name" type="string"/>
      <property name="user_email" column="user_email" type="string"/>
      <property name="password" column="password" type="string"/>
      <property name="created" type="timestamp" column="created"/>
   </class>
</hibernate-mapping>

tweet.hbm

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE hibernate-mapping PUBLIC 
 "-//Hibernate/Hibernate Mapping DTD//EN"
 "hibernate-mapping-3.0.dtd">
<hibernate-mapping>
   <class name="com.csc.project.tweet" table="t2">
      <meta attribute="class-description">
         This class contains the employee detail. 
      </meta>
      <id name="tweet_id" type="int" column="tweet_id">
         <generator class="increment"/>
      </id>


     <property name="user_id" column="user_id" type="int"/>
      <property name="message" column="message" type="string"/>
     <!--  <property name="user_email" column="user_email" type="string"/>
      <property name="password" column="password" type="string"/> -->
      <property name="created" type="timestamp" column="created"/>
   </class>
</hibernate-mapping>

Answer 1

我认为你的bean属性是DB字段的名称

如果您想获得关于一个用户的推文，您的HQL将是：

select e.user_name, p.message 
from tweet p, t.com.csc.project.user e
where p.user_id=e.id
and e.user_name=:name

在字段之前使用始终别名，您可以使用隐式连接（使用逗号）替换显式连接

如何在hql中使用内连接查询？

1 个答案: