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[英]HQL Left Join - org.hibernate.hql.internal.ast.ErrorCounter - Path expected for join
[英]HQL left join: Path expected for join
我是Hibernate的新手,我有一個關於HQL左連接的問題。
我嘗試離開加入2個表,患者和提供者,並繼續獲得“預期加入的路徑!” 第二個表上的錯誤。 如果有人可以幫助解決這個問題,請欣賞它!
這是2個表/類的映射:
patient.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>
provider.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
}
}
這是左連接查詢:
select
pat.patientId, pat.patientName
from
patient as pat
left join
provider as pro
where
pat.providerId = pro.providerId
這是結果:
0:50:08,479 INFO query:156 - Query = outerJoin 10:50:08,479 INFO query:157 - select pat.patientId, pat.patientName from patient as pat left join provider as pro where pat.providerId = pro.providerId 10:50:08,698 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 null 10:50:08,698 ERROR query:184 - Problem generating query. org.hibernate.hql.ast.QuerySyntaxException: Path expected for join! [select pat.patientId, pat.patientName from com.ccg.db.test.patient as pat left join provider as pro where pat.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)
您的患者可以參考提供者,並且還將提供者ID作為財產。 我可能會擺脫患者的提供者ID屬性,只需要提供者的參考。 然后你的查詢應該是這樣的。
select pat.patientId, pat.patientName
from patient as pat
left join pat.provider as pro
要加入,您需要從患者到提供者的關聯路徑,在本例中為pat.provider。 然后,hibernate將自動使用多對一映射中指定的“列”加入提供程序表。 在你的情況下,連接沒有多大意義,因為你沒有查詢提供者的任何屬性,所以這樣的事情可能更有意義
select pat
from patient as pat
join pat.provider as pro
where pat.patientName = 'John'
and pro.name = 'United Healthcare'
在那里,您可以將患者名單過濾給名為John的患者,這些患者將United Healthcare作為提供者。
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