Question

我一直得到一个＆＃34;无效的标识符＆＃34;在我输入此代码时DepartmentName：

SELECT LastName "Last Name", Gender "Gender"
FROM Employee
WHERE DepartmentName =
(SELECT DepartmentName FROM Department WHERE DepartmentName = 'Radiology');

感谢任何帮助！

Answer 1

如果在创建表时你的列名中有双引号（说“DepartmentName”），那么你也应该在select query中有双引号的列。

如果您使用Where DepartmentName，您将收到错误ora-00904-invalid-identifier

因此，为避免错误，您必须使用WHERE“DepartmentName”

Answer 2

我猜这个意图是这样的：

SELECT e.LastName as "Last Name", e.Gender as "Gender"
FROM Employee e
WHERE e.DepartmentId = (SELECT d.DepartmentId
                        FROM Department d
                        WHERE d.DepartmentName = 'Radiology'
                       );

也就是说，您希望在某种id上连接这两个表。如果DepartmentName中有Employee，您只需写：

SELECT e.LastName as "Last Name", e.Gender as "Gender"
FROM Employee e
WHERE e..DepartmentName = 'Radiology';

不需要“加入”。

Answer 3

部门的名称可能不存储在员工表中，但有些＆＃34;外键＆＃34;在Department表中的一行是。因此，您将加入这两个表

SELECT e.LastName as "Last Name", e.Gender as "Gender"
FROM Employee e
INNER JOIN Department d ON e.DepartmentId = d.id -- guessed this column name
WHERE d.DepartmentName = 'Radiology'

如果遇到连接或IN（）的选择，请使用连接。

查询中的标识符无效

3 个答案: