具有多对多关系的多个表中的 JPA 查询

Question

共有三个表：Hospital、Medical_Service和Language_Service， 医院可提供医疗服务和语言服务。所以有两个多对多关系。

Simple ERD

现在我想用postcode = 3000 和medical service = Emergency 搜索医院数据。

DaoImpl：

public List<Hospital> findByPostcodeAndMedicalType(String postcode, String medical) {
        String str = "SELECT h FROM Hospital h INNER JOIN Medical_Service m ON h.hospital_id = m.hospital_id WHERE " 
                + "h.Postcode = :postcode AND m.Medical_name = :medical";
        Query query = em.createQuery(str);
        query.setParameter("postcode", postcode);
        query.setParameter("medical", medical);
        return query.getResultList();

    }

另外，如果我想从三个表中按邮政编码、医疗类型和语言进行搜索，如何写一个jsql。

警告：

错误：org.hibernate.hql.internal.ast.ErrorCounter - 预期加入的路径！ 预计加入的路径！ 在 org.hibernate.hql.internal.ast.HqlSqlWalker.createFromJoinElement(HqlSqlWalker.java:378) 在 org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.joinElement(HqlSqlBaseWalker.java:3858) 在 org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.fromElement(HqlSqlBaseWalker.java:3644)

2016 年 4 月 2 日晚上 10:54:30 org.apache.catalina.core.StandardWrapperValve 调用 严重：servlet [appServlet] 在路径 [/travel] 的上下文中的 Servlet.service() 引发异常 [请求处理失败；嵌套异常是 java.lang.IllegalArgumentException: org.hibernate.QueryException: could not resolve property: Postcode of: com.health.entity.Hospital [SELECT h FROM com.health.entity.Hospital h INNER加入 Medical_Service m ON h.hospital_id = m.hospital_id WHERE h.Postcode = :postcode AND m.Medical_name = :medical]] 与根本原因 org.hibernate.QueryException：无法解析属性：邮政编码：com.health.entity.Hospital 在 org.hibernate.persister.entity.AbstractPropertyMapping.propertyException（AbstractPropertyMapping.java:83） 在 org.hibernate.persister.entity.AbstractPropertyMapping.toType(AbstractPropertyMapping.java:77) 在 org.hibernate.persister.entity.AbstractEntityPersister.toType(AbstractEntityPersister.java:1978) 在 org.hibernate.hql.internal.ast.tree.FromElementType.getPropertyType(FromElementType.java:367)

Hospital.class

@Entity
@Table(name = "Hospital")
public class Hospital {

@Id
@GeneratedValue
    private int hospital_id;

    private String hospital_name;

    private String postcode;

    private String suburb;

    private String address;

    private String type;

    private String category;

    private String longitude;

    private String latitude;

    private String email;

    private String website;

    private String phoneno;

    private String isemergency;

    private String agencytype;

    private String fax;

    @ManyToMany
    @JoinTable(
        name = "Hospital_Medical",
        joinColumns=@JoinColumn(name="Hospital_id", referencedColumnName="Hospital_id"),
        inverseJoinColumns=@JoinColumn(name="Medical_id", referencedColumnName="Medical_id"))
private List<MedicalService> services;

    @ManyToMany
    @JoinTable(
        name = "Hospital_Language",
        joinColumns=@JoinColumn(name="Hospital_id", referencedColumnName="Hospital_id"),
        inverseJoinColumns=@JoinColumn(name="Language_id", referencedColumnName="Language_id"))
private List<Language> languages;

//Setter and Getter
}

MedicalService.class

@Entity
@Table(name = "Medical_Service")
public class MedicalService {

@Id
private int medical_id;

private String medical_name;

private String description;

@ManyToMany(mappedBy="services")
private List<Hospital> hospitals;
//Setter and Getter
}

Language.class

@Entity
@Table(name = "Language")
public class Language {

@Id
private int language_id;

private String language_name;

private String display_name;

@ManyToMany(mappedBy="languages")
private List<Hospital> hospitals;
//Setter and Getter
}

Answer 1

我认为您的查询可能是错误的，这可能是问题的原因。

您目前正在使用：

 SELECT h FROM Hospital h
     INNER JOIN Medical_Service m ON h.hospital_id = m.hospital_id
     WHERE h.Postcode = :postcode AND m.Medical_name = :medical

问题可能是 Medical_Service 不包含 Hospital_id 字段（在 JOIN 中使用）。

如果您乐于使用本机查询，您可以这样做：

 SELECT * FROM Hospital WHERE Postcode = 3000 AND Hospital_id IN
    (SELECT Hospital_id FROM Hospital_Medical hm INNER JOIN Medical_Service m ON hm.Medical_id = m.Medical_id
    where Medical_name = 'Emergency')

内部 SELECT 获取提供紧急服务的医院的所有 Hospital_id。然后，外部选择会选择 Hospital_id 在内部 SELECT 中的所有医院（即它们提供紧急服务），但也只选择邮政编码为 3000 的医院。

要使用本机查询，您需要执行以下操作：

    int postcode = 3000;
    String service = "Emergency";

    StringBuilder sb = new StringBuilder(); 
        sb.append("SELECT * FROM Hospital WHERE Postcode = ");
        sb.append(postcode);
        sb.append("AND Hospital_id IN SELECT Hospital_id FROM Hospital_Medical hm INNER JOIN "
                + "Medical_Service m ON hm.Medical_id = m.Medical_id where Medical_name = '");
        sb.append(service);
        sb.append("')");

    String queryString = sb.toString();
    Query query = em.createNativeQuery(queryString);
    List<Hospital> result = query.getResultList();