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select dept_name,avg_salary from ( select dept_name, avg(salary) as avg_salary from instructor groupby 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 groupby 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 inset