未识别的名称：[9：8]的员工

发布于 2025-01-26 07:49:05 字数 536 浏览 2 评论 0原文

这给我带来了未识别的名称错误。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

 未识别的名称：[9：8]的员工

在此处输入图像描述

原文

This is giving me an unrecognized name error. Why?

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

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云之铃。 2025-02-02 07:49:05

您缺少别名员工和部门完整表引用！您在子句中使用它们，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

you are missing aliases employees and departments after full table references! you use them in ON clause but you missed to define them!

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

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早乙女 2025-02-02 07:49:05

您需要给桌子别名。我会建议表名的缩写：

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;

You need to give your tables aliases. I would recommend abbreviations for the table names:

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;

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