java - HQL 左连接 : Path expected for join

Question

我是 Hibernate 的新手，我有一个关于 HQL Left join 的问题。

我尝试左连接 2 个表，患者和提供者，并不断收到“Path expected for join!”第二张表上的错误。如果有人可以帮助解决这个问题，我们将不胜感激!

这是 2 个表/类的映射:

患者.hbm.xml:

<?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 package="com.ccg.db.test">
    <class name="patient" table="patient">
        <id name="patientId" column="patientId" type="int">
            <generator class="native"/>
        </id>
        <property name="patientName" type="string" >
           <column name="patientName" /> 
        </property>
        <property name="providerId" type="string" >
            <column name="provId" /> 
        </property>
        <many-to-one name="provider" column="providerId" class="provider" /> 
    </class>
</hibernate-mapping>

---

提供者.hbm.xml:

<?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 package="com.ccg.db.test">
    <class name="provider" table="provider">
        <id name="providerId" column="providerId">
            <generator class="native" />
        </id>
        <property name="providerName" column="providerName" />
    </class>
</hibernate-mapping>

---

POJO:

patient.java

package com.ccg.db.test;

import java.io.Serializable;
import java.util.List;
import org.hibernate.Session;

public class patient 
implements Serializable
{
    private int patientId;
    private String patientName;
    private String providerId; // foreign key to provider

    private static final long serialVersionUID = 81073;

    public static void load(Session session, List<String> values){
        patient PatientInfo = new patient(); 

        PatientInfo.setPatientId(Integer.parseInt(values.get(0)));
        PatientInfo.setPatientName( values.get(1));
        PatientInfo.setProviderId( values.get(2) );

        session.save( PatientInfo );
    }

    /**
    * @return the PatientId
    */
    public int getPatientId() {
        return patientId;
    }

    /**
    * @param PatientId the PatientId to set
    */
    public void setPatientId(int PatientId) {
        this.patientId = PatientId;
    }

    /**
    * @return the PatientName
    */
    public String getPatientName() {
        return this.patientName;
    }

    /**
    * @param PatientName the PatientName to set
    */
    public void setPatientName(String PatientName) {
        this.patientName = PatientName;
    }

    /**
    * @return the ProvId
    */
    public String getProviderId() {
        return this.providerId;
    }

    /**
    * @param id the ProviderId to set
    */
    public void setProviderId( String id ) {
        this.providerId = id;
    }

    /**
    * @return the ProvId
    */
    public String getProvider() {
        return this.providerId;
    }

    /**
    * @param id the ProviderId to set
    */
    public void setProvider( String id ) {
        this.providerId = id;
    }  

    /**
    * @param args
    */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
    }
}

provider.java:

package com.ccg.db.test;

import java.io.Serializable;
import java.util.List;

import org.hibernate.Session;

public class provider 
implements Serializable
{
    private String providerId;
    private String providerName;

    //private int patientId;
    //private int providerSpec; 

    private static final long serialVersionUID = 81073;

    public static void load(Session session, List<String> values){
        provider ProviderInfo = new provider(); 

        ProviderInfo.setProviderId( values.get(0) );
        ProviderInfo.setProviderName( values.get(1));
        //ProviderInfo.setProviderSpec( Integer.parseInt(values.get(2)) );

        session.save( ProviderInfo );
    }

    /**
    * @return the ProviderName
    */
    public String getProviderName() {
        return providerName;
    }

    /**
    * @param ProviderName the ProviderName to set
    */
    public void setProviderName(String name) {
        this.providerName = name;
    }

    /**
    * @return the ProvId
    */
    public String getProviderId() {
        return this.providerId;
    }

    /**
    * @param id the ProvId to set
    */
    public void setProviderId( String id ) {
        this.providerId = id;
    }

    /*
    public int getPatientId() {
        return this.patientId;
    }

    public void setPatientId( int id ) {
        this.patientId = id;
    }
    */

    /**
    * @param args
    */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
    }
}

---

这是左连接查询:

select
    pat.patientId, pat.patientName
from
    patient as pat 
left join
    provider as pro 
where
    pat.providerId = pro.providerId

---

结果如下:

0:50:08,479 INFO query:156 - Query = outerJoin10:50:08,479 INFO query:157 - selectpat.patientId, pat.patientNamefrompatient as pat left joinprovider as pro wherepat.providerId = pro.providerId10:50:08,698 ERROR PARSER:33 - Path expected for join!10:50:08,698 ERROR PARSER:33 - Invalid path: 'pro.providerId'10:50:08,698 ERROR PARSER:33 - right-hand operand of a binary operator was null10:50:08,698 ERROR query:184 - Problem generating query.org.hibernate.hql.ast.QuerySyntaxException: Path expected for join! [selectpat.patientId, pat.patientNamefromcom.ccg.db.test.patient as pat left joinprovider as pro wherepat.providerId = pro.providerId]at org.hibernate.hql.ast.QuerySyntaxException.convert(QuerySyntaxException.java:31)at org.hibernate.hql.ast.QuerySyntaxException.convert(QuerySyntaxException.java:24)at org.hibernate.hql.ast.ErrorCounter.throwQueryException(ErrorCounter.java:59)at org.hibernate.hql.ast.QueryTranslatorImpl.analyze(QueryTranslatorImpl.java:235)at org.hibernate.hql.ast.QueryTranslatorImpl.doCompile(QueryTranslatorImpl.java:160)at org.hibernate.hql.ast.QueryTranslatorImpl.compile(QueryTranslatorImpl.java:111)at org.hibernate.engine.query.HQLQueryPlan.(HQLQueryPlan.java:77)at org.hibernate.engine.query.HQLQueryPlan.(HQLQueryPlan.java:56)at org.hibernate.engine.query.QueryPlanCache.getHQLQueryPlan(QueryPlanCache.java:72)at org.hibernate.impl.AbstractSessionImpl.getHQLQueryPlan(AbstractSessionImpl.java:133)at org.hibernate.impl.AbstractSessionImpl.createQuery(AbstractSessionImpl.java:112)at org.hibernate.impl.SessionImpl.createQuery(SessionImpl.java:1623)at com.ccg.db.query.QueryManager.query(QueryManager.java:163)at com.ccg.db.query.QueryManager.query(QueryManager.java:139)at com.ccg.db.query.QueryManager.main(QueryManager.java:80)

Answer 1

您的患者有对提供者的引用，并且还具有提供者 ID 作为属性。我可能会去掉患者的提供者 ID 属性，只引用提供者。那么您的查询应该是这样的。

select pat.patientId, pat.patientName 
from patient as pat 
left join pat.provider as pro

要加入，您需要从患者到提供者的关联路径，在本例中为 pat.provider。然后 hibernate 将自动使用多对一映射中指定的“列”加入提供者表。在你的情况下，加入没有多大意义，因为你没有查询提供者的任何属性，所以这样的事情可能更有意义

select pat 
from patient as pat 
join pat.provider as pro 
where pat.patientName = 'John' 
and pro.name = 'United Healthcare'

在那里，您可以将您的患者列表过滤为以 United Healthcare 作为提供者的名为 John 的患者。