Question

这是我的jpaContext.xml：

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xmlns:tx="http://www.springframework.org/schema/tx"
    xmlns:context="http://www.springframework.org/schema/context"
    xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd
        http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.2.xsd
        http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx-3.2.xsd">
    <context:annotation-config />
    <context:component-scan base-package="com.pluralsight"/>

    <bean
        class="org.springframework.orm.jpa.support.PersistenceAnnotationBeanPostProcessor">
    </bean>
    <bean id="entityManagerFactory"
        class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
    <property name="persistenceUnitName" value="punit"></property>
    <property name="dataSource" ref="dataSource"></property>
    <property name="jpaVendorAdapter">
        <bean
            class="org.springframework.orm.jpa.vendor.HibernateJpaVendorAdapter">
        <property name="showSql" value="true"></property>
        </bean>
    </property>
    <property name="jpaPropertyMap">
        <map>
            <entry key="hibernate.dialect"
                value="org.hibernate.dialect.MySQL5InnoDBDialect">
            </entry>
            <entry key="hibernate.hbm2ddl.auto" value="none"></entry>
            <entry key="hibernate.format_sql" value="true"></entry>
        </map>
    </property>
    </bean>
    <bean id="transactionManager"
        class="org.springframework.orm.jpa.JpaTransactionManager">
        <property name="entityManagerFactory"
            ref="entityManagerFactory">
        </property>
    </bean>
    <tx:annotation-driven transaction-manager="transactionManager" />
    <bean id="dataSource"
        class="org.springframework.jdbc.datasource.DriverManagerDataSource">
        <property name="driverClassName"
            value="com.microsoft.sqlserver.jdbc.SQLServerDriver">
        </property>
        <property name="url"
            value="jdbc:sqlserver://123.123.123.123:1433;databaseName=WikiGenome">
        </property>
        <property name="username" value="xxx"></property>
        <property name="password" value="xxx"></property>
    </bean>
</beans>

这是我的Disease.java：

@Entity
@Table(name="Disease")
public class Disease {
  @Id
  @GeneratedValue
  @Column(name="DiseaseID")
  public int DiseaseID;

  @Column(name="Name")
  public String Name;
}

这是我的另一个课程：

@Entity
@Table(name="ChrPosDisease")
public class ChrPosDisease implements Serializable{
    @Id
    @Column(name="chr")
    public String chr;

    @Id
    @Column(name="pos")
    public int pos;

    @Id
    @Column(name="DiseaseID")
    public int diseaseID;
}

我是hibernate和spring mvc框架的新手，我只是按照教程中的指南进行操作。我可以使用：

查询结果

@SuppressWarnings({ "unchecked"})
    public List getDiseaseByName(String name) {
        Query query = em.createQuery("Select d From Disease d Where d.Name=?1").setParameter(1,name);
        List diseaseList=query.getResultList();
        return diseaseList;
    }

但是，当我通过DiseaseID连接两个表时，它会出现以下错误。

java.lang.IllegalArgumentException: org.hibernate.hql.internal.ast.QuerySyntaxException: Path expected for join! [Select d From cuhk.cse.fyp.model.Disease d JOIN ChrPosDisease c Where d.DiseaseID=c.DiseaseID AND d.Name=?1]

仅使用一个表查询结果时，我没有收到错误。

以下是我用来加入这两个表的代码：

@SuppressWarnings({ "unchecked"})
    public List getJoinDiseaseByName(String name) {
        Query query = em.createQuery("Select d From Disease d JOIN ChrPosDisease c Where d.DiseaseID=c.DiseaseID AND d.Name=?1").setParameter(1,name);
        List diseaseList=query.getResultList();
        return diseaseList;
    }

怎么了？

感谢您的帮助。

补充： 更新了ChrPosDisease

@SuppressWarnings("serial")
@Entity
@Table(name="ChrPosDisease")
public class ChrPosDisease implements Serializable{
    @Id
    @Column(name="chr")
    public String chr;

    @Id
    @Column(name="pos")
    public int pos;

    @Column(name="DiseaseID")
    public int diseaseID;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name="DiseaseID",nullable=false)
    private Disease disease;
}

更新疾病：

@SuppressWarnings("serial")
@Entity
@Table(name="Disease")
public class Disease implements Serializable{
    @Id
    @GeneratedValue
    @Column(name="DiseaseID")
    public int DiseaseID;

    @Column(name="Name")
    public String Name;

    @OneToMany(fetch = FetchType.LAZY, mappedBy = "ChrPosDisease")
    private Set<ChrPosDisease> chrPosDisease = new HashSet<ChrPosDisease>();
}

我使用了上面的实体，部署时有例外。

我还需要添加其他内容吗？

Answer 1

我认为你不应该使用d.ChrPosDisease而只使用ChrPosDisease，因为d只是疾病实体的别名。

它应该有用。

让我知道它是否无效。

Answer 2

您应该将映射条目设为：

@OneToMany(fetch = FetchType.LAZY, mappedBy = "disease")
private Set<ChrPosDisease> chrPosDisease = new HashSet<ChrPosDisease>();

mappedBy属性通知该字段由该特定实体属性映射。因此，此属性应该是ManyToOne侧连接实体的属性。

此处Disease有许多ChrPosDisease。 ChrPosDisease有一个Disease。因此，按列映射的应该是OneToMany字段与ManyToOne侧的属性绑定的字段。

无法构建EntityManagerFactory - 表连接中出错

2 个答案: