使用 INSERT INTO .. SELECT * FROM 时，INSERT into 语句在 oracle 中不起作用答案

【问题标题】：INSERT into statement not working in oracle when using INSERT INTO .. SELECT * FROM使用 INSERT INTO .. SELECT * FROM 时，INSERT into 语句在 oracle 中不起作用
【发布时间】：2019-10-26 13:31:38
【问题描述】：

我有vh_emp 表，其表结构是（这是空表）：

我正在从 employee 表中获取数据作为源表，我在其中编写了一些查询以匹配 vh_emp 表：

SELECT e.emp_id       EMP_ID, 
       e.emp_name     EMP_NAME, 
       e.dob          DOB, 
       e.join_date    JOIN_DATE, 
       CASE 
         WHEN e.leave_date IS NULL THEN Trunc(sysdate) 
         ELSE e.leave_date 
       END            AS LEAVE_DATE, 
       e.salary / 100 SALARY, 
       d.dept_name    DEPT_NAME, 
       one.addr_name  ADDRESS_ONE, 
       two.addr_name  ADDRESS_TWO, 
       b.emp_name     AS MANAGER_NAME 
FROM   offc.employee e 
       LEFT JOIN offc.department d 
              ON e.dept_id = d.dept_id 
       LEFT JOIN offc.add_line_one one 
              ON e.line1 = one.addr_id_one 
       LEFT JOIN offc.add_line_two two 
              ON e.line2 = two.addr_id_two 
       LEFT OUTER JOIN (SELECT * 
                        FROM   offc.employee) b 
                    ON e.manager_id = b.emp_id;

查询运行良好，该查询的输出也正常：

所以，我想将这些数据插入到vh_emp 表中：

INSERT INTO offc.vh_emp 
            (emp_id, 
             emp_name, 
             dob, 
             join_date, 
             leave_date, 
             address_one, 
             address_two, 
             salary, 
             manager_name, 
             dept_name) 
SELECT e.emp_id       EMP_ID, 
       e.emp_name     EMP_NAME, 
       e.dob          DOB, 
       e.join_date    JOIN_DATE, 
       CASE 
         WHEN e.leave_date IS NULL THEN Trunc(sysdate) 
         ELSE e.leave_date 
       END            AS LEAVE_DATE, 
       e.salary / 100 SALARY, 
       d.dept_name    DEPT_NAME, 
       one.addr_name  ADDRESS_ONE, 
       two.addr_name  ADDRESS_TWO, 
       b.emp_name     AS MANAGER_NAME 
FROM   offc.employee e 
       LEFT JOIN offc.department d 
              ON e.dept_id = d.dept_id 
       LEFT JOIN offc.add_line_one one 
              ON e.line1 = one.addr_id_one 
       LEFT JOIN offc.add_line_two two 
              ON e.line2 = two.addr_id_two 
       LEFT OUTER JOIN (SELECT * 
                        FROM   offc.employee) b 
                    ON e.manager_id = b.emp_id;

但是，我执行了这个查询我得到了错误：

ORA-01722: invalid number

表employee的结构为：

为什么会出现这个错误？我查看并看到 select 语句的列和值的数据类型相同。由于列数据类型匹配，因此不应出现此错误。

【问题讨论】：

标签： sql oracle oracle11g sql-insert

【解决方案1】：

insert 子句列名和select 子句列名的顺序应该匹配。

请按如下方式匹配他们的位置：

INSERT INTO offc.vh_emp 
            (emp_id, 
             emp_name, 
             dob, 
             join_date, 
             leave_date, 
             Salary,  -- this
             dept_name, -- this
             address_one, 
             address_two,
             manager_name 
             ) 
SELECT e.emp_id       EMP_ID, 
       e.emp_name     EMP_NAME, 
       e.dob          DOB, 
       e.join_date    JOIN_DATE, 
       CASE 
         WHEN e.leave_date IS NULL THEN Trunc(sysdate) 
         ELSE e.leave_date 
       END            AS LEAVE_DATE, 
       e.salary / 100 SALARY, 
       d.dept_name    DEPT_NAME, 
       one.addr_name  ADDRESS_ONE, 
       two.addr_name  ADDRESS_TWO, 
       b.emp_name     AS MANAGER_NAME 
FROM   offc.employee e 
       LEFT JOIN offc.department d 
              ON e.dept_id = d.dept_id 
       LEFT JOIN offc.add_line_one one 
              ON e.line1 = one.addr_id_one 
       LEFT JOIN offc.add_line_two two 
              ON e.line2 = two.addr_id_two 
       LEFT OUTER JOIN (SELECT * 
                        FROM   offc.employee) b 
                    ON e.manager_id = b.emp_id;

干杯！！

【讨论】：