Question

我有这个表：员工（id，名字，姓氏，工资），我添加了另一列，用于存储每位员工过去的所有工资。

CREATE OR REPLACE TYPE salary_list AS TABLE OF NUMBER;
/
ALTER TABLE employees
ADD (salary_history salary_list)
NESTED TABLE salary_history STORE AS salary_history_tab;

现在，该表是员工（id，firstname，lastname，salary， salary_history ），salary_history为null。当我修改salaray值时，如何将当前工资值插入内部表salary_history？我试过：

INSERT INTO 
TABLE(SELECT salary_history FROM employees WHERE id=1) 
VALUES(1500);

我收到以下错误：

ORA-22908：对NULL表值的引用

是的，嵌套表salary_history为null，因为我刚刚创建它。我究竟做错了什么？将值（附加到现有数据）插入 salary_history 以正确识别员工的正确方法是什么？

更新：像这样添加了nvl（，（））：

INSERT INTO 
TABLE(SELECT nvl(salary_history,salary_list()) FROM employees WHERE id=1) 
VALUES(1000);

我现在得到的错误是：

ORA-25015：无法在此嵌套表视图列
上执行DML

Answer 1

您可以使用COALESCE( salary_history, salary_list() ) MULTISET UNION ALL salary_list( :your_new_value )将新值附加到旧列表中（如果不存在则创建新列表）。

Oracle 11g R2架构设置：

CREATE OR REPLACE TYPE salary_list AS TABLE OF NUMBER;
/

CREATE TABLE employees(
  id             NUMBER,
  salary_history salary_list
) NESTED TABLE salary_history STORE AS salary_history_tab
/

INSERT INTO employees VALUES ( 1, NULL )
/

查询1 ：

UPDATE employees
SET   salary_history = COALESCE( salary_history, salary_list() )
                       MULTISET UNION ALL salary_list( 500 )
WHERE id = 1

SELECT * FROM employees

<强> Results ：

| ID | SALARY_HISTORY |
|----|----------------|
|  1 |            500 |

查询2 ：

UPDATE employees
SET   salary_history = COALESCE( salary_history, salary_list() )
                       MULTISET UNION ALL salary_list( 700 )
WHERE id = 1

SELECT * FROM employees

<强> Results ：

| ID | SALARY_HISTORY |
|----|----------------|
|  1 |        500,700 |

查询3 ：

UPDATE employees
SET   salary_history = COALESCE( salary_history, salary_list() )
                       MULTISET UNION ALL salary_list( 500 )
WHERE id = 1

SELECT * FROM employees

<强> Results ：

| ID | SALARY_HISTORY |
|----|----------------|
|  1 |    500,700,500 |

Answer 2

该列当前为null，而不是空表;您可以创建一个空的嵌套表来开始：

UPDATE employees
SET salary_history = NEW salary_list()
WHERE id = 1;

1 row updated.

然后你的第一个陈述将起作用：

INSERT INTO 
TABLE(SELECT salary_history FROM employees WHERE id=1) 
VALUES(1500);

1 row inserted.

您可以看到新值：

SELECT * FROM employees;

        ID FIRSTNAME  LASTNAME       SALARY SALARY_HISTORY                
---------- ---------- ---------- ---------- ------------------------------
         1 Joe        BLoggs           1234 SALARY_LIST(1500)

创建嵌套表并仅在内部表

2 个答案: