java错误组织。冬眠例外SQLGrammarException:无法提取结果集

Question

我有一个正在执行的hql查询，但收到了一个错误，我并不真正理解其原因

这是我的代码：

@Override
public List<StaffRequest> getStaffLeaveRequest(String userID, Date startDate, Date endDate) 
{
    Session currentSession = sessionFactory.getCurrentSession();        

    List<StaffRequest> results = 
    currentSession.createQuery("select new com.timesheet_Webservice.CustomEntity.StaffRequest(lr.leave_ID, lr.leave_Employee, concat(s.staff_First_Name, ' ', s.staff_Last_Name), "
            + "(lr.leave_Days*8.5), lr.leave_Comments, '1805', concat(pro.project_Pastel_Prefix, ' - ', pro.project_Description), lr.leave_Start, lr.leave_End, lr.leave_IsApproved, "
            + "(select lt.leaveType_Description from LeaveType lt where lt.leaveType_ID = lr.leave_Type)) "
            + "from Staff s, Project pro, Leave lr "
            + "where lr.leave_Employee = s.staff_Code and pro.project_Code = 1805 and lr.leave_Approved = :userID and lr.leave_IsApproved = 0 and s.staff_IsEmployee <> 0 "
            + "and lr.leave_Start between :startDate and :endDate "
            + "order by concat(s.staff_First_Name, ' ', s.staff_Last_Name)")
            .setParameter("userID",userID).setParameter("startDate", startDate).setParameter("endDate", endDate).getResultList();   

    return results;
}

在尝试执行时，我在网页上出现以下错误：

org.hibernate.exception.SQLGrammarException: could not extract ResultSet

还有这个控制台错误：

ERROR: You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'Leave leave2_ where leave2_.Leave_Employee=staff0_.Staff_Code and project1_.Proj' at line 1

这似乎表明where子句有错误，但我看不出有什么特别的错误

更新： 实体类

项目

@Entity
@Table(name="project")
public class Project {

    @Id
    @GeneratedValue(strategy=GenerationType.IDENTITY)
    @Column(name="Project_Code")
    public int project_Code;

    @Column(name="Project_Customer")
    public int project_Customer;

    //A lot more attributes...

}

员工

@Entity
@Table(name="staff")
public class Staff 
{
    @Id
    @Column(name="Staff_Code")
    public String staff_Code;
    ...
}

Answer 1

显然，JPA遵循变量名的某些规则或惯例，不接受“ux”，并且出于某种原因不接受中间大写字母。。。它通过将所有变量都改为小写来对我起作用

Answer 2

好的，经过多次测试，我发现了错误的原因。由于某种未知的原因，该查询似乎与被引用表“leave”的名称有问题，每当我试图从中检索数据时，都会产生错误。但是，如果我将表重命名为像“leaves”这样简单的名称，那么查询将成功执行。任何人都可能知道这是为什么