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解决MySQL错误 Every derived table must have its own alias

发表于 更新于 分类于 ， ， Valine：


问题描述

今天使用如下的子查询时:

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select dept_name,avg_salary
from ( select dept_name, avg(salary) as avg_salary
    from instructor 
    group by dept_name
)
where avg_salary>42000;

报了:Every derived table must have its own alias这个错误
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mysql> select dept_name,avg_salary
from ( select dept_name, avg(salary) as avg_salary
    from instructor 
    group by dept_name
)
where avg_salary>42000;
1248 - Every derived table must have its own alias

解决方案

Every derived table must have its own alias翻译成中文就是:每个派生表都必须有自己的别名
所以给子查询取一个别名即可:

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select dept_name,avg_salary
from ( select dept_name, avg(salary) as avg_salary
    from instructor 
    group by dept_name
) as I
where avg_salary>42000;

这样就能正常执行了:
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mysql> select dept_name,avg_salary
from ( select dept_name, avg(salary) as avg_salary
    from instructor 
    group by dept_name
) as I
where avg_salary>42000;
+------------+--------------+
| dept_name  | avg_salary   |
+------------+--------------+
| Biology    | 72000        |
| Comp. Sci. | 77333.333333 |
| Elec. Eng. | 80000        |
| Finance    | 85000        |
| History    | 61000        |
| Physics    | 91000        |
+------------+--------------+
6 rows in set