为什么要获得期望加入的org.hibernate.hql.internal.ast.QuerySyntaxException:路径！连接3个表时

Question

我有三个实体basicInfo，性别和部门。如果我加入了basicInfo和性别模型，那么它可以正常工作，但是当我想和其他两个实体同时加入部门时，就会显示出来。

.QuerySyntaxException:期望加入的路径！

我的三个实体类，

@Entity
@Table(name = "employeebasicinfo")
public class BasicInfoModel {
@Id private String employeeId;
@NotBlank private String firstName;
@NotBlank private String lastName;
@NotNull private Integer companyId;
@Transient private String companyName;
private String phoneNo;

@Column(name = "sex")
private char sexId;

@Transient private String sexName;

@Temporal(TemporalType.TIMESTAMP)
private Date birthDate;

@Temporal(TemporalType.TIMESTAMP)
private Date joiningDate;

@NotNull private Integer department;
@Transient private String departmentName;

@ManyToOne()
@JoinColumn(name = "sex", referencedColumnName = "id", insertable = false, updatable = false)
private GenderModel genderModel;

@ManyToOne
@JoinColumn(
  name = "department",
  referencedColumnName = "id",
  insertable = false,
  updatable = false)
private DepartmentModel departmentModel;
// getter setter
 }

性别模型

 @Entity
 @Table(name = "gender")
 public class GenderModel implements Serializable {
 @Id
 @Column(name = "id")
 private char id;

 @Column(name = "name")
 private String name;
 //getter setter
}

与部门模型

   @Entity
   @Table(name = "department")
   public class DepartmentModel {
   @Id private int id;
   private String name;
   //getter setter
}

我的Dto类

public class BasicInfoDto {
  private String employeeId;
  private String firstName;
  private String lastName;
  private Integer companyId;
  private String companyName;
  private String phoneNo;
  private char sexId;
  private String sexName;
  private Date birthDate;
  private Date joiningDate;
  private Integer department;
  private String departmentName;


  public BasicInfoDto(char sexId, String sexName, int department, String departmentName) {
    this.sexId = sexId;
    this.sexName = sexName;
    this.department = department;
    this.departmentName = departmentName;

   //getter setter
  }

我的存储库类如下所示：

@Repository
public interface BasicInfoRepository extends JpaRepository<BasicInfoModel, String> {


  @Query(
      value =
          "SELECT  new com.employeeAttendance.employeeAttendance.dto.BasicInfoDto ("
              + " a.sexId, b.name, a.department,c.name) FROM "
              + " BasicInfoModel a INNER JOIN GenderModel b on a.sexId = b.id" +
                   " INNER JOIN DepartmentMode c on a.department = c.id")
  List<BasicInfoDto> getBasicInfoList();
}

，最后，我得到了这个错误

由: org.hibernate.hql.internal.ast.QuerySyntaxException:路径引起的: java.lang.IllegalArgumentException:期望加入！选择新的com.employeeAttendance.employeeAttendance.dto.BasicInfoDto ( a.sexId，b.name，a.department，从com.employeeAttendance.employeeAttendance.model.BasicInfoModel a内部连接com.employeeAttendance.employeeAttendance.model.GenderModel b on a.sexId = b.id内连接DepartmentMode c on a.department = c.id at org.hibernate.internal.ExceptionConverterImpl.convert(ExceptionConverterImpl.java:138) ~hibernate-5.3.7.final.jar:5.3.7 INNER at org.hibernate.internal.ExceptionConverterImpl.convert(ExceptionConverterImpl.java:181) ~hibernate-core-5org.hibernate.internal.ExceptionConverterImpl.convert(ExceptionConverterImpl.java:188) ~hibernate处的.3.7.Final.jar:5.3.7.Final -core-5.3.7.Final.jar:5.3.7.Final.jar在org.hibernate.internal.AbstractSharedSessionContract.createQuery(AbstractSharedSessionContract.java:713) ~hibernate-core-5.3.7.Final.jar:5.3.7.Final.jar在org.hibernate.internal.AbstractSessionImpl.createQuery(AbstractSessionImpl.java:23) ~hibernate-core-5.3.7.final.jar:5.3.7.Final.jar: sun.reflect.NativeMethodAccessorImpl.invoke0(Native方法完成) ~na:1.8.0_191 at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62) ~na:1.8.0_191 at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43) ~na:1.8.0_191 at java.lang.reflect.Methodorg.springframework.orm.jpa.ExtendedEntityManagerCreator$ExtendedEntityManagerInvocationHandler.invoke(ExtendedEntityManagerCreator.java:350) ~ .invoke(Method.java:498) ~na:1.8.0_191 at .invoke~spring-orm-5.1.4 RELEASE.jar:5.1.4.RELEASE at com.sun.agent。$Proxy109.createQuery(未知来源) ~na:na at org.springframework.data.jpa.repository.query.SimpleJpaQuery.validateQuery(SimpleJpaQuery.java:87) ~spring-data-jpa-2。1.4.RELEASE.jar:2.1.4.RELEASE .由于: org.hibernate.hql.internal.ast.QuerySyntaxException:路径预期会加入而忽略了76个常见帧！选择新的com.employeeAttendance.employeeAttendance.dto.BasicInfoDto ( a.sexId，b.name，a.department，从com.employeeAttendance.employeeAttendance.model.BasicInfoModel a内部连接com.employeeAttendance.employeeAttendance.model.GenderModel b on a.sexId = b.id内部连接DepartmentMode c on a.department = c.id at org.hibernate.hql.internal.ast.QuerySyntaxException.convert(QuerySyntaxException.java:74) ~hibernate-核心-5.3.7.Final.jar:5.3.7#en3# at org.hibernate.hql.internal.ast.ErrorTracker.throwQueryException(ErrorTracker.java：93) ~hibernate-core-5.3.7.final.jar:5.3.7.Final.jar在org.hibernate.hql.internal.ast.QueryTranslatorImpl.analyze(QueryTranslatorImpl.java:277) ~hibernate-core-5.3.7.final.jar:5.3.7.final.jar:5.3.7.Final.jar在org.hibernate.hql.internal.ast.QueryTranslatorImpl.doCompile(QueryTranslatorImpl.java:191) ~hibernate-核心-5.3.7.Final.jar:5.3.7.final.jar:5.3.7.final.jar在org.hibernate.hql.internal.ast.QueryTranslatorImpl.compile(QueryTranslatorImpl.java:143) ~hibernate-core- 5.3.7.final.jar:5.3.7.Final.jar在org.hibernate.engine.query.spi.HQLQueryPlan.(HQLQueryPlan.java:119) ~hibernate-核心-5.3.7.Final.jar:5.3.7.Final.jar在org.hibernate.engine.query.spi.HQLQueryPlan.(HQLQueryPlan.java:80) ~hibernate-core5.3.7.Final.jar:5.3.7.Final在org.hibernate.engine.query.spi.QueryPlanCache.getHQLQueryPlan(QueryPlanCache.java:153) ~hibernate-core-5.3.7.Final.jar:5.3.7.Final.jar在org.hibernate.internal.AbstractSharedSessionContract.getQueryPlan(AbstractSharedSessionContract.java:595) ~hibernate-core-5.3.7.Final.jar:5.3.7在org.hibernate.internal.AbstractSharedSessionContract.createQuery(AbstractSharedSessionContract.java:704) ~hibernate-core-5.3.7.Final.jar:5.3.7.Final.jar.84个常见帧被省略

但是，如果我删除了第二次与部门的连接，那么它就能正常工作。

Answer 1

您正在处理jpql中的实体属性，而不是表。因此，您只告诉jpql要使用哪个属性来连接：

"SELECT  new com.employeeAttendance.employeeAttendance.dto.BasicInfoDto ("
  + " a.sexId, b.name, a.department,c.name) FROM "
  + " BasicInfoModel a INNER JOIN a.genderModel b "
  + " INNER JOIN a.departmentModel c";