Question

我有三个实体basicInfo，性别和部门。如果我加入basicInfo和性别模型，则可以正常工作，但是当要同时与其他两个实体一起加入部门时，则显示

.QuerySyntaxException：预期要加入的路径！

我的三个实体在这里

@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
}

和部门模型

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

我的Dto课程

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
  }

我的存储库类别如下：

@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();
}

最后我收到此错误

由于：java.lang.IllegalArgumentException： org.hibernate.hql.internal.ast.QuerySyntaxException：预期的路径加入！ [选择新 com.employeeAttendance.employeeAttendance.dto.BasicInfoDto（a.sexId， b.name，a.department，c.name）FROM com.employeeAttendance.employeeAttendance.model.BasicInfoModel内部加入com.employeeAttendance.employeeAttendance.model.GenderModel b a.sexId = b.id内部联接a。部门上的DepartmentMode c = c.id] 在org.hibernate.internal.ExceptionConverterImpl.convert（ExceptionConverterImpl.java:138）〜[hibernate-core-5.3.7.Final.jar：5.3.7.Final] 在org.hibernate.internal.ExceptionConverterImpl.convert（ExceptionConverterImpl.java:181）〜[hibernate-core-5.3.7.Final.jar：5.3.7.Final] 在org.hibernate.internal.ExceptionConverterImpl.convert（ExceptionConverterImpl.java:188）〜[hibernate-core-5.3.7.Final.jar：5.3.7.Final] 在org.hibernate.internal.AbstractSharedSessionContract.createQuery（AbstractSharedSessionContract.java:713）〜[hibernate-core-5.3.7.Final.jar：5.3.7.Final] 在org.hibernate.internal.AbstractSessionImpl.createQuery（AbstractSessionImpl.java:23）〜[hibernate-core-5.3.7.Final.jar：5.3.7.Final] 在sun.reflect.NativeMethodAccessorImpl.invoke0（本机方法）〜[na：1.8.0_191] 在sun.reflect.NativeMethodAccessorImpl.invoke（NativeMethodAccessorImpl.java:62）〜[na：1.8.0_191] 在sun.reflect.DelegatingMethodAccessorImpl.invoke（DelegatingMethodAccessorImpl.java:43）〜[na：1.8.0_191] 在java.lang.reflect.Method.invoke（Method.java:498）〜[na：1.8.0_191] 在org.springframework.orm.jpa.ExtendedEntityManagerCreator $ ExtendedEntityManagerInvocationHandler.invoke（ExtendedEntityManagerCreator.java:350）处〜[spring-orm-5.1.4.RELEASE.jar：5.1.4.RELEASE] 在com.sun.proxy。$ Proxy109.createQuery（未知来源）〜[na：na] 在org.springframework.data.jpa.repository.query.SimpleJpaQuery.validateQuery（SimpleJpaQuery.java:87）〜[spring-data-jpa-2.1.4.RELEASE.jar：2.1.4.RELEASE] ...省略了76个通用框架引起原因：org.hibernate.hql.internal.ast.QuerySyntaxException：预期加入的路径！ [选择新 com.employeeAttendance.employeeAttendance.dto.BasicInfoDto（a.sexId， b.name，a.department，c.name）FROM com.employeeAttendance.employeeAttendance.model.BasicInfoModel内部加入com.employeeAttendance.employeeAttendance.model.GenderModel b a.sexId = b.id内部联接a。部门上的DepartmentMode c = c.id] 在org.hibernate.hql.internal.ast.QuerySyntaxException.convert（QuerySyntaxException.java:74）〜[hibernate-core-5.3.7.Final.jar：5.3.7.Final] 在org.hibernate.hql.internal.ast.ErrorTracker.throwQueryException（ErrorTracker.java:93）〜[hibernate-core-5.3.7.Final.jar：5.3.7.Final] 在org.hibernate.hql.internal.ast.QueryTranslatorImpl.analyze（QueryTranslatorImpl.java:277）〜[hibernate-core-5.3.7.Final.jar：5.3.7.Final] 在org.hibernate.hql.internal.ast.QueryTranslatorImpl.doCompile（QueryTranslatorImpl.java:191）〜[hibernate-core-5.3.7.Final.jar：5.3.7.Final] 在org.hibernate.hql.internal.ast.QueryTranslatorImpl.compile（QueryTranslatorImpl.java:143）〜[hibernate-core-5.3.7.Final.jar：5.3.7.Final] 在org.hibernate.engine.query.spi.HQLQueryPlan。（HQLQueryPlan.java:119）〜[hibernate-core-5.3.7.Final.jar：5.3.7.Final] 在org.hibernate.engine.query.spi.HQLQueryPlan。（HQLQueryPlan.java:80）〜[hibernate-core-5.3.7.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] 在org.hibernate.internal.AbstractSharedSessionContract.getQueryPlan（AbstractSharedSessionContract.java:595）〜[hibernate-core-5.3.7.Final.jar：5.3.7.Final] 在org.hibernate.internal.AbstractSharedSessionContract.createQuery（AbstractSharedSessionContract.java:704）〜[hibernate-core-5.3.7.Final.jar：5.3.7.Final] ...省略了84个通用框架

但是，如果我删除了部门的第二次加入，那么它可以正常工作。

Answer 1

您正在使用jpql中的实体属性，而不是在表上。因此，您只告诉jpql使用哪个属性来加入：

"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";

为什么要获得org.hibernate.hql.internal.ast.QuerySyntaxException：期望加入的路径！加入3张桌子时

1 个答案: