无法识别的姓名:[9:8] 的员工
Unrecognized name: employees at [9:8]
这是一个无法识别的名称错误。为什么?
SELECT
employees.name AS employee_name,
employees.role AS employee_role,
departments.name AS department_name
FROM
`strange-calling-318804.employee_data.Employees`
JOIN
`strange-calling-318804.employee_data.departments`
ON employees.department_id = departments.department_id
Unrecognized name: employees at [9:8]
enter image description here
您需要提供 table 的别名。我会推荐 table 名称的缩写:
SELECT e.name AS employee_name, e.role AS employee_role,
d.name AS department_name
FROM `strange-calling-318804.employee_data.Employees` e JOIN
`strange-calling-318804.employee_data.departments` d
ON e.department_id = d.department_id;
在完整的 table 引用之后,您缺少别名 employees
和 departments
!您在 ON 子句中使用了它们,但您没有定义它们!
SELECT
employees.name AS employee_name,
employees.role AS employee_role,
departments.name AS department_name
FROM
`strange-calling-318804.employee_data.Employees` as employees
JOIN
`strange-calling-318804.employee_data.departments` as departments
ON employees.department_id = departments.department_id
非常感谢。
coursera 数据分析课程非常好,但有些视频缺少一些更新,这就是其中之一。他们没有提到我们必须在 ON 语法上输入别名。
在不修复此错误的情况下尝试观看视频让我很头疼。
这是一个无法识别的名称错误。为什么?
SELECT
employees.name AS employee_name,
employees.role AS employee_role,
departments.name AS department_name
FROM
`strange-calling-318804.employee_data.Employees`
JOIN
`strange-calling-318804.employee_data.departments`
ON employees.department_id = departments.department_id
Unrecognized name: employees at [9:8]
enter image description here
您需要提供 table 的别名。我会推荐 table 名称的缩写:
SELECT e.name AS employee_name, e.role AS employee_role,
d.name AS department_name
FROM `strange-calling-318804.employee_data.Employees` e JOIN
`strange-calling-318804.employee_data.departments` d
ON e.department_id = d.department_id;
在完整的 table 引用之后,您缺少别名 employees
和 departments
!您在 ON 子句中使用了它们,但您没有定义它们!
SELECT
employees.name AS employee_name,
employees.role AS employee_role,
departments.name AS department_name
FROM
`strange-calling-318804.employee_data.Employees` as employees
JOIN
`strange-calling-318804.employee_data.departments` as departments
ON employees.department_id = departments.department_id
非常感谢。
coursera 数据分析课程非常好,但有些视频缺少一些更新,这就是其中之一。他们没有提到我们必须在 ON 语法上输入别名。
在不修复此错误的情况下尝试观看视频让我很头疼。