Sql 使用INSERT into.时，INSERT into语句在oracle中不起作用。。从中选择*

我有

vh_emp

表，其表结构为（这是空表）：

我将

employee

表中的数据作为源表，在源表中我编写了一些查询以与vh_emp表匹配：

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; 

查询运行正常，此查询的输出正常：

因此，我想使用以下方法在

vh_emp

表格中插入这些数据：

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，得到的错误如下：

ORA-01722: invalid number

表

employee

的结构是：

---

为什么会出现这种错误？我查看并发现select语句的列和值的数据类型相同。当列数据类型匹配时，不应出现此错误。

插入子句列名和选择子句列名的顺序应匹配

请将他们的位置匹配如下：

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

插入子句列名和选择子句列名的顺序应匹配

请将他们的位置匹配如下：

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