将 SQL 查询转换为 JPQL - @ManyToMany 加入
Translate SQL Query to JPQL - @ManyToMany join
我正在尝试将 SQL @Query("...", native=true) Spring Data Jpa 注释中的本机查询用法转换为同一注释中的 JPQL 查询用法.
SQL 查询：
select d.id as doctorId, d.firstName as doctorFirstName, d.lastName as doctorLastName, d.title as doctorTitle,
d.email as doctorEmail, v.id as id, v.dateFrom as dateFrom, v.dateTo as dateTo, v.status as status,
md.id as medicalServicesId, md.service as medicalServicesService, md.price as medicalServicesPrice
from Visit v
left outer join Doctor d on v.doctor_id=d.id
left outer join visit_medical_services vms on v.id=vms.medical_services_id
left outer join MedicalService md on vms.visit_id=md.id
where d.id= :doctorId and v.status= :status and v.dateFrom>= :dateFrom and v.dateTo<= :dateTo
我正在使用 Spring Projected Interface，这就是列别名如此命名的原因。
现在我想使用 JPQL - 项目列表获得完全相同的结果。我试过的是这样的：
@Query("select d.id as doctorId, d.firstName as doctorFirstName, d.lastName as doctorLastName, d.title as doctorTitle, d.email as doctorEmail, " +
"v.id as id, v.dateFrom as dateFrom, v.dateTo as dateTo, v.status as status, md.id as medicalServicesId, md.service as medicalServicesService, md.price as medicalServicesPrice \n" +
"from Visit v \n" +
"left outer join Doctor d on v.doctor.id=d.id \n" +
"left outer join v.medicalServices vms on vms.id=v.id \n" +
"left outer join MedicalService md on md.id=vms.id \n" +
"where d.id= :doctorId and v.status= :status and v.dateFrom>= :dateFrom and v.dateTo<= :dateTo")
List<VisitInfoWithPatientAndMedServices3joins> getAllVisitInfoWithPatientAndMedicalServicesJpqlQuery
(@Param("doctorId") Long doctorId, @Param("status") VisitStatus status, @Param("dateFrom") LocalDateTime dateFrom, @Param("dateTo") LocalDateTime dateTo, Pageable pageable);
此处翻译自 JPQL SQL。
select doctor1_.id as col_0_0_, doctor1_.firstName as col_1_0_, doctor1_.lastName as col_2_0_, doctor1_.title as col_3_0_, doctor1_.email as col_4_0_, visit0_.id as col_5_0_, visit0_.dateFrom as col_6_0_, visit0_.dateTo as col_7_0_, visit0_.status as col_8_0_, medicalser4_.id as col_9_0_, medicalser4_.service as col_10_0_, medicalser4_.price as col_11_0_
from Visit visit0_
left outer join
(visit_medical_services medicalser2_ left outer join MedicalService medicalser3_ on medicalser2_.visit_id=medicalser3_.id) on visit0_.id=medicalser2_.medical_services_id
and (medicalser3_.id=visit0_.id)
left outer join Doctor doctor1_ on (visit0_.doctor_id=doctor1_.id)
left outer join MedicalService medicalser4_ on (medicalser4_.id=medicalser3_.id) where doctor1_.id=? and visit0_.status=? and visit0_.dateFrom>=? and visit0_.dateTo<=? limit ?
这里的问题是，它从@ManyToMany 关系 (visit_medical_services) 返回列，但仅针对第一个对象。 Json 以下邮递员的回复：
[
{
"id": 1,
"status": "PAID",
"dateFrom": "2019-04-10T08:00:00",
"dateTo": "2019-04-10T08:30:00",
"medicalServicesId": 1,
"medicalServicesService": "Visit",
"medicalServicesPrice": 100.0,
"doctorLastName": "James",
"doctorFirstName": "Alex",
"doctorEmail": "james@gmail.com",
"doctorId": 1,
"doctorTitle": "dr n. md."
},
{
"id": 2,
"status": "PAID",
"dateFrom": "2019-04-10T09:00:00",
"dateTo": "2019-04-10T09:30:00",
"medicalServicesId": null,
"medicalServicesService": null,
"medicalServicesPrice": null,
"doctorLastName": "James",
"doctorFirstName": "Alex",
"doctorEmail": "james@gmail.com",
"doctorId": 1,
"doctorTitle": "dr n. md."
}
]
我尝试在 Workbench 中使用使用 JPQL 的已翻译 SQL 查询，因为我使用的是 MySQL 数据库并且结果是相同的 - 第一个对象是正确的，并且其余在映射的@ManyToMany 列中具有空值。
这是我的实体 类 如果它能让这个问题更容易一些：
public class Visit {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
Long id;
LocalDateTime dateFrom;
LocalDateTime dateTo;
@Enumerated(EnumType.STRING)
VisitStatus status;
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "doctor_id", referencedColumnName = "id", nullable = true)
@JsonManagedReference
Doctor doctor;
@ManyToOne(fetch = FetchType.LAZY)
@JsonManagedReference
@JoinColumn(name = "patient_id", referencedColumnName = "id", nullable = true)
Patient patient;
@ManyToMany
@JsonManagedReference
@JoinTable(
name = "visit_diseases",
joinColumns = @JoinColumn(
name = "disease_id", referencedColumnName = "id"),
inverseJoinColumns = @JoinColumn(
name = "visit_id", referencedColumnName = "id"))
List<Disease> diseases;
@ManyToMany
@JsonManagedReference
@JoinTable(
name = "visit_medical_services",
joinColumns = @JoinColumn(
name = "medical_services_id"),
inverseJoinColumns = @JoinColumn(
name = "visit_id"))
Set<MedicalService> medicalServices;
String mainSymptoms;
String treatment;
String allergy;
String addiction;
String comment;
}
和医疗服务：
public class MedicalService {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "id")
Long id;
String service;
Float price;
@ManyToMany(mappedBy = "medicalServices", fetch = FetchType.EAGER)
private Set<Visit> visits;
public MedicalService(Long id, String service, float price) {
this.id = id;
this.service = service;
this.price = price;
}
}
有人可以看一下并向我解释这里有什么问题吗？我想要实现的是让 JPQL 生成 THE SAME SQL 查询。这甚至可能吗？请帮助我...
您的查询有一些错误。请改用以下内容
@Query("select d.id as doctorId, d.firstName as doctorFirstName, d.lastName as doctorLastName, d.title as doctorTitle, d.email as doctorEmail, " +
"v.id as id, v.dateFrom as dateFrom, v.dateTo as dateTo, v.status as status, md.id as medicalServicesId, md.service as medicalServicesService, md.price as medicalServicesPrice \n" +
"from Visit v \n" +
"left outer join v.doctor d \n" +
"left outer join v.medicalServices md \n"
"where d.id= :doctorId and v.status= :status and v.dateFrom>= :dateFrom and v.dateTo<= :dateTo")
List<VisitInfoWithPatientAndMedServices3joins> getAllVisitInfoWithPatientAndMedicalServicesJpqlQuery
(@Param("doctorId") Long doctorId, @Param("status") VisitStatus status, @Param("dateFrom") LocalDateTime dateFrom, @Param("dateTo") LocalDateTime dateTo, Pageable pageable);
我正在尝试将 SQL @Query("...", native=true) Spring Data Jpa 注释中的本机查询用法转换为同一注释中的 JPQL 查询用法. SQL 查询：
select d.id as doctorId, d.firstName as doctorFirstName, d.lastName as doctorLastName, d.title as doctorTitle,
d.email as doctorEmail, v.id as id, v.dateFrom as dateFrom, v.dateTo as dateTo, v.status as status,
md.id as medicalServicesId, md.service as medicalServicesService, md.price as medicalServicesPrice
from Visit v
left outer join Doctor d on v.doctor_id=d.id
left outer join visit_medical_services vms on v.id=vms.medical_services_id
left outer join MedicalService md on vms.visit_id=md.id
where d.id= :doctorId and v.status= :status and v.dateFrom>= :dateFrom and v.dateTo<= :dateTo
我正在使用 Spring Projected Interface，这就是列别名如此命名的原因。
现在我想使用 JPQL - 项目列表获得完全相同的结果。我试过的是这样的：
@Query("select d.id as doctorId, d.firstName as doctorFirstName, d.lastName as doctorLastName, d.title as doctorTitle, d.email as doctorEmail, " +
"v.id as id, v.dateFrom as dateFrom, v.dateTo as dateTo, v.status as status, md.id as medicalServicesId, md.service as medicalServicesService, md.price as medicalServicesPrice \n" +
"from Visit v \n" +
"left outer join Doctor d on v.doctor.id=d.id \n" +
"left outer join v.medicalServices vms on vms.id=v.id \n" +
"left outer join MedicalService md on md.id=vms.id \n" +
"where d.id= :doctorId and v.status= :status and v.dateFrom>= :dateFrom and v.dateTo<= :dateTo")
List<VisitInfoWithPatientAndMedServices3joins> getAllVisitInfoWithPatientAndMedicalServicesJpqlQuery
(@Param("doctorId") Long doctorId, @Param("status") VisitStatus status, @Param("dateFrom") LocalDateTime dateFrom, @Param("dateTo") LocalDateTime dateTo, Pageable pageable);
此处翻译自 JPQL SQL。
select doctor1_.id as col_0_0_, doctor1_.firstName as col_1_0_, doctor1_.lastName as col_2_0_, doctor1_.title as col_3_0_, doctor1_.email as col_4_0_, visit0_.id as col_5_0_, visit0_.dateFrom as col_6_0_, visit0_.dateTo as col_7_0_, visit0_.status as col_8_0_, medicalser4_.id as col_9_0_, medicalser4_.service as col_10_0_, medicalser4_.price as col_11_0_
from Visit visit0_
left outer join
(visit_medical_services medicalser2_ left outer join MedicalService medicalser3_ on medicalser2_.visit_id=medicalser3_.id) on visit0_.id=medicalser2_.medical_services_id
and (medicalser3_.id=visit0_.id)
left outer join Doctor doctor1_ on (visit0_.doctor_id=doctor1_.id)
left outer join MedicalService medicalser4_ on (medicalser4_.id=medicalser3_.id) where doctor1_.id=? and visit0_.status=? and visit0_.dateFrom>=? and visit0_.dateTo<=? limit ?
这里的问题是，它从@ManyToMany 关系 (visit_medical_services) 返回列，但仅针对第一个对象。 Json 以下邮递员的回复：
[
{
"id": 1,
"status": "PAID",
"dateFrom": "2019-04-10T08:00:00",
"dateTo": "2019-04-10T08:30:00",
"medicalServicesId": 1,
"medicalServicesService": "Visit",
"medicalServicesPrice": 100.0,
"doctorLastName": "James",
"doctorFirstName": "Alex",
"doctorEmail": "james@gmail.com",
"doctorId": 1,
"doctorTitle": "dr n. md."
},
{
"id": 2,
"status": "PAID",
"dateFrom": "2019-04-10T09:00:00",
"dateTo": "2019-04-10T09:30:00",
"medicalServicesId": null,
"medicalServicesService": null,
"medicalServicesPrice": null,
"doctorLastName": "James",
"doctorFirstName": "Alex",
"doctorEmail": "james@gmail.com",
"doctorId": 1,
"doctorTitle": "dr n. md."
}
]
我尝试在 Workbench 中使用使用 JPQL 的已翻译 SQL 查询，因为我使用的是 MySQL 数据库并且结果是相同的 - 第一个对象是正确的，并且其余在映射的@ManyToMany 列中具有空值。 这是我的实体 类 如果它能让这个问题更容易一些：
public class Visit {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
Long id;
LocalDateTime dateFrom;
LocalDateTime dateTo;
@Enumerated(EnumType.STRING)
VisitStatus status;
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "doctor_id", referencedColumnName = "id", nullable = true)
@JsonManagedReference
Doctor doctor;
@ManyToOne(fetch = FetchType.LAZY)
@JsonManagedReference
@JoinColumn(name = "patient_id", referencedColumnName = "id", nullable = true)
Patient patient;
@ManyToMany
@JsonManagedReference
@JoinTable(
name = "visit_diseases",
joinColumns = @JoinColumn(
name = "disease_id", referencedColumnName = "id"),
inverseJoinColumns = @JoinColumn(
name = "visit_id", referencedColumnName = "id"))
List<Disease> diseases;
@ManyToMany
@JsonManagedReference
@JoinTable(
name = "visit_medical_services",
joinColumns = @JoinColumn(
name = "medical_services_id"),
inverseJoinColumns = @JoinColumn(
name = "visit_id"))
Set<MedicalService> medicalServices;
String mainSymptoms;
String treatment;
String allergy;
String addiction;
String comment;
}
和医疗服务：
public class MedicalService {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "id")
Long id;
String service;
Float price;
@ManyToMany(mappedBy = "medicalServices", fetch = FetchType.EAGER)
private Set<Visit> visits;
public MedicalService(Long id, String service, float price) {
this.id = id;
this.service = service;
this.price = price;
}
}
有人可以看一下并向我解释这里有什么问题吗？我想要实现的是让 JPQL 生成 THE SAME SQL 查询。这甚至可能吗？请帮助我...
您的查询有一些错误。请改用以下内容
@Query("select d.id as doctorId, d.firstName as doctorFirstName, d.lastName as doctorLastName, d.title as doctorTitle, d.email as doctorEmail, " +
"v.id as id, v.dateFrom as dateFrom, v.dateTo as dateTo, v.status as status, md.id as medicalServicesId, md.service as medicalServicesService, md.price as medicalServicesPrice \n" +
"from Visit v \n" +
"left outer join v.doctor d \n" +
"left outer join v.medicalServices md \n"
"where d.id= :doctorId and v.status= :status and v.dateFrom>= :dateFrom and v.dateTo<= :dateTo")
List<VisitInfoWithPatientAndMedServices3joins> getAllVisitInfoWithPatientAndMedicalServicesJpqlQuery
(@Param("doctorId") Long doctorId, @Param("status") VisitStatus status, @Param("dateFrom") LocalDateTime dateFrom, @Param("dateTo") LocalDateTime dateTo, Pageable pageable);