0

这给了我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]

在此处输入图像描述

4

3 回答 3

1

你需要给你的表别名。我会推荐表名的缩写:

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;
于 2021-08-23T20:52:12.547 回答
1

您缺少别名employees并且departments在完整的表引用之后!您在 ON 子句中使用它们,但您错过了定义它们!

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
于 2021-08-23T20:53:11.017 回答
0

太感谢了。

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 ALIASES。

试图在不修复此错误的情况下观看视频让我很头疼。

于 2022-02-03T01:13:43.217 回答