jpa复合主键表不返回值

Posted 2023-04-14

【中文标题】jpa复合主键表不返回值

【英文标题】：jpa composite primary key table not returning values

【发布时间】：2014-01-31 08:19:08

【问题描述】：

我是新 JPA，当我从复合主键表中检索值时遇到异常。

异常说明：

Problem compiling [select t from ASSIGN_TASK_EMPLOYEE t]. 
    [14, 34] The abstract schema type 'ASSIGN_TASK_EMPLOYEE' is unknown.
        at org.eclipse.persistence.internal.jpa.EntityManagerImpl.createQuery(EntityManagerImpl.java:1605)

以下是我的代码

@Entity
        @Table(name = "ASSIGN_TASK_EMPLOYEE")
        //@IdClass(AssignTaskEmployeePk.class)
        public class AssignTaskEmployee implements Serializable 

            @EmbeddedId
            private AssignTaskEmployeePk assignTaskEmployeePk;

            public AssignTaskEmployeePk getAssignTaskEmployeePk() 
                return assignTaskEmployeePk;
            

            public void setAssignTaskEmployeePk(AssignTaskEmployeePk assignTaskEmployeePk) 
                this.assignTaskEmployeePk = assignTaskEmployeePk;
            

        

       @Embeddable
        public class AssignTaskEmployeePk  

            private String employeeId;
            private String taskId;
            public AssignTaskEmployeePk() 
            

            @Override
            public boolean equals(Object obj) 
                // TODO Auto-generated method stub

                if (obj instanceof AssignTaskEmployeePk) 

                    AssignTaskEmployeePk employeePk = (AssignTaskEmployeePk) obj;
                    if (!employeePk.getEmployeeId().equals(this.employeeId)) 
                        return false;
                               
                    else if (!employeePk.getTaskId().equals(this.taskId)) 
                        return false;
                    

                
                else 
                    return false;
                

                return false;

            

            @Override
            public int hashCode()       
                return employeeId.hashCode() + taskId.hashCode() ;
            

            public String getEmployeeId() 
                return employeeId;
            

            public void setEmployeeId(String employeeId) 
                this.employeeId = employeeId;
            

            public String getTaskId() 
                return taskId;
            

            public void setTaskId(String taskId) 
                this.taskId = taskId;
            


            

我在数据库中为组合主键 ASSIGN_TASK_EMPLOYEE（表）添加了四个值，其中 PK 表

EMP_ID   TASKID
1         2
2         4
3         5
4         6 

现在我想获得分配给 emp_id 1 的任务 为此，我编写了以下查询：这应该返回 AssignTaskEmployee 对象的列表。

entityManager.createQuery("select t from ASSIGN_TASK_EMPLOYEE t").getResultList()

当我执行这个查询时，我得到以下异常

Exception Description: 

Problem compiling [select t from ASSIGN_TASK_EMPLOYEE t]. 
            [14, 34] The abstract schema type 'ASSIGN_TASK_EMPLOYEE' is unknown.
                at org.eclipse.persistence.internal.jpa.EntityManagerImpl.createQuery(EntityManagerImpl.java:1605)

【参考方案1】：

JPQL 应该使用实体名称，默认是类的名称。分配任务员工

应该是

entityManager.createQuery("select t from AssignTaskEmployee  t").getResultList()

以上将返回表 ASSIGN_TASK_EMPLOYEE 中的所有记录。

如果您想使用 JPQL 检索特定记录，您应该使用 WHERE 语句，如下所示：

    Query query =  entityManager.createQuery("select t from AssignTaskEmployee t WHERE 
t.assignTaskEmployeePk.employeeId = :employeeId and t.assignTaskEmployeePk.taskId = :taskId")

query.setParameter("employeeId", 1);
query.setParameter("taskId",1);

query.getSingleResult()  //As expected to have only one record.

阅读this查询EmbeddedId