这给了我一个无法识别的名字错误。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无法识别的姓名: 9:8的雇员
发布于 2021-08-23 20:53:11
在完整的表引用之后,您缺少别名employees和departments!您在ON子句中使用它们,但是您忽略了对它们的定义!
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发布于 2021-08-23 20:52:12
你得给你的桌子取个别名。我建议表名的缩写如下:
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;https://stackoverflow.com/questions/68898855
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