Question

我正在使用一个带有确定性用户定义函数的虚拟列（oracle 11g），该函数将该行的主键作为参数并返回一个简单的标量值。虚拟列更新没有任何问题，但是当我更新表时，它会抛出错误： - ora-00054资源忙，并在oracle 指定nowait或超时时获取。我的表结构和功能如下： -

 -----------------------------------------------------------------------
   id   employee_name    employee_dept  employee_leaves (vir column)
 -----------------------------------------------------------------------
   2     patrick           mgmt         getEmpLeaves(id)  
   3      jack             sales            "
 -----------------------------------------------------------------------

     create or replace function getEmpLeaves(empId number) 
        return number 
          DETERMINISTIC
       is
           emp_leaves number;
       begin
           select leaves into emp_leaves from tbl_emp_leaves 
           where tbl_emp_leaves.id = empId;
           return emp_leaves;
      end ;
    -------------------------------------------------------------

如何克服此错误？

Answer 1

我不打算找出这个错误的原因简短的回答是：从表中删除此虚拟列，然后创建一个视图：

create view vw_employees AS
SELECT t.id, t.employee_name, t.employee_dept, x.leaves As employee_leaves
FROM tbl_employees t
JOIN tbl_emp_leaves x
ON t.id = x.id;

答案很长：请看下面简单的测试用例：

create table tbl_emp_leaves as
select object_id as id, trunc(dbms_random.value(0,100)) as leaves
from all_objects;
alter table tbl_emp_leaves add primary key( id );

create or replace function getEmpLeaves(empId number) 
        return number 
          DETERMINISTIC
       is
           emp_leaves number;
       begin
           select leaves into emp_leaves from tbl_emp_leaves 
           where tbl_emp_leaves.id = empId;
           return emp_leaves;
      end ;
      /

create table tbl_employees as
select object_id as id, object_name as employee_name, object_type as employee_dept
from all_objects;

alter table tbl_employees 
add employee_leaves as ( getEmpLeaves(id)); 

create view vw_employees AS
SELECT t.id, t.employee_name, t.employee_dept, x.leaves As employee_leaves
FROM tbl_employees t
JOIN tbl_emp_leaves x
ON t.id = x.id;

现在比较两个简单查询的性能：

SQL> set timing on;
SQL> select sum(employee_leaves) from vw_employees;

SUM(EMPLOYEE_LEAVES)
--------------------
             3675425

Elapsed: 00:00:00.07
SQL> select sum(employee_leaves) from tbl_employees;

SUM(EMPLOYEE_LEAVES)
--------------------
             3675425

Elapsed: 00:00:03.09

3.09秒vs. 0.07秒 - 您会发现基于该功能的虚拟列比简单连接慢44倍（即：4400％）。

Answer 2

当我创建一个选择另一个表的函数时，我遇到了类似的问题。

关键字DETERMINISTIC表示函数始终为给定的输入参数返回相同的值。开发人员有责任确保这一点。

Oracle允许这样的函数：

create or replace function getNumber(x number) 
    return number DETERMINISTIC is
begin
  return DBMS_RANDOM.RANDOM;
end;

虽然这与 deterministic 函数完全相反。

在UPDATE tbl_emp_leaves SET leaves = ...语句后，函数的返回值会发生变化。

所以，我必须同意kordirko的回答：删除此虚拟列并创建视图或使用触发器设置列employee_leaves的值。

使用确定性用户定义函数的虚拟列

2 个答案: