将值插入表Oracle SQL

Question

我正在尝试将值插入Oracle SQL中的"Employee"表.我有一个关于输入由外键确定的值的问题:

我的员工有3个由外键确定的属性:State,Position和Manager.我使用INSERT INTO语句插入值并手动键入数据.我是否需要在物理上查找每个引用以输入数据,或者是否有可以使用的命令？例如

INSERT INTO Employee 
(emp_id, emp_name, emp_address, emp_state, emp_position, emp_manager)
VALUES 
(001, "John Doe", "1 River Walk, Green Street", 3, 5, 1000)

这应该填充employee表(John Doe, 1 River Walk, Green Street, New York, Sales Executive, Barry Green).纽约state_id=3在State桌子上; 销售主管position_id=5在positions谈判桌上; 和巴里格林manager_id=1000在manager桌子上.

有没有办法可以输入引用表的文本值,以便Oracle识别文本并将其与相关ID匹配？我希望这个问题有意义,很乐意澄清任何事情.

谢谢!

Answer 1

您可以使用以下函数,以便在插入之前从DB中提取更多参数:

--
-- insert_employee  (Function) 
--
CREATE OR REPLACE FUNCTION insert_employee(p_emp_id in number, p_emp_name in varchar2, p_emp_address in varchar2, p_emp_state in varchar2, p_emp_position in varchar2, p_emp_manager in varchar2) 
RETURN VARCHAR2 AS

   p_state_id varchar2(30) := '';
 BEGIN    
      select state_id 
      into   p_state_id
      from states where lower(emp_state) = state_name;

      INSERT INTO Employee (emp_id, emp_name, emp_address, emp_state, emp_position, emp_manager) VALUES 
                (p_emp_id, p_emp_name, p_emp_address, p_state_id, p_emp_position, p_emp_manager);

    return 'SUCCESS';

 EXCEPTION 
   WHEN others THEN
    RETURN 'FAIL';
 END;
/

Answer 2

INSERT
INTO    Employee 
        (emp_id, emp_name, emp_address, emp_state, emp_position, emp_manager)
SELECT  '001', 'John Doe', '1 River Walk, Green Street', state_id, position_id, manager_id
FROM    dual
JOIN    state s
ON      s.state_name = 'New York'
JOIN    positions p
ON      p.position_name = 'Sales Executive'
JOIN    manager m
ON      m.manager_name = 'Barry Green'

请注意,但是单个拼写错误(或额外空格)将导致不匹配,并且不会插入任何内容.