Question

有三个表：Hospital，Medical_Service和Language_Service，医院可以提供医疗服务和语言服务。所以有两对多关系。

现在，我想使用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();

    }

此外，如果我想从三个表中搜索邮政编码，医疗类型和语言，如何编写jsql。

警告：

错误：org.hibernate.hql.internal.ast.ErrorCounter - 加入的路径！    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日下午10:54:30 org.apache.catalina.core.StandardWrapperValve调用     严重：Servlet [appServlet]的Servlet.service（）在路径[/ travel]的上下文中引发异常[请求处理失败;嵌套异常是java.lang.IllegalArgumentException：org.hibernate.QueryException：无法解析属性：postcode ：com.health.entity.Hospital [SELECT h FROM com.health.entity.Hospital h INNER JOIN Medical_Service m ON h.hospital_id = m.hospital_id WHERE h.Postcode =：postcode AND m.Medical_name =：medical]]有根本原因     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）         at 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有误，这可能是问题的原因。

您目前正在使用：

#include <IE.au3>
$oIE = _IECreate("http://www.google.com")
$oAs = _IETagnameGetCollection($oIE, "input")
$i = 1
For $oA In $oAs
   If $i = 4 Then _IEPropertySet($oA, "innertext", "MyValue")
   $i = $i + 1
Next

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

如果您乐意使用原生查询，可以执行以下操作：

 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

内部SELECT获取提供紧急服务的医院的所有Hospital_id。然后外部选择选择Hospital_id在内部SELECT中的所有医院（即他们提供紧急服务），但也选择那些邮政编码为3000的医院。

要使用原生查询，您可以执行以下操作：

 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')

具有多对多关系的多个表中的JPA查询

1 个答案: