Solution | get the relevant information about the current employees' highest salary in each department |

Source of problem :
sql Difficult questions :SQL12: Get information about the highest paid employees in each department

difficulty : Solve and search the aggregation process ,sql need select Non aggregate Columns
Solutions : Apply two sub tables to query , An aggregation result for the aggregation column ( Non aggregate column ), One contains Unaggregated results of aggregated columns and Non aggregate Columns

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***Mysql And Sqlite select Non aggregate Columns *** The result is Not the same , If this question only requires output :

dept_no,maxSalary Two , Only use max Aggregate functions :

select de.dept_no,max(sa.salary) as maxSalary
from dept_emp as de left join salaries as sa
on de.emp_no=sa.emp_no
group by de.dept_no
order by de.dept_no asc

But ontology needs select emp_no This non aggregated field , And an aggregate field (dept_no) Corresponding to multiple non aggregated fields (emp_no), So when you choose ,Sqlite Any one of the non aggregated fields will be randomly selected ( Not choice max(sa.salary) The one corresponding to the maximum salary emp_no), and mysql Will report an error directly !
This is a select Non aggregate fields , Classical problems !!!

resolvent :
First construct a sub table t1( The salary value of the Contract Department ), Used to find the maximum salary max(sa.salary) With its dept_no,

(
    select de.dept_no,max(sa.salary) as salary
    from dept_emp as de left join salaries as sa
    on de.emp_no=sa.emp_no
    group by dept_no
) as t1

Then build a sub table t2( Do not aggregate the salary value of the same department in the table , Merge only two tables ), Storage dept_no,salary(t2 Medium dept_no,salary For and t1 Medium dept_no,salary comparison ),emp_no

(
    select de.dept_no,de.emp_no,sa.salary
    from dept_emp as de left join salaries as sa
    on de.emp_no=sa.emp_no
) as t2

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t1,t2 Joint query of two sub tables :

select t2.dept_no,t2.emp_no,t1.salary as maxSalary from
(
    select de.dept_no,max(sa.salary) as salary
    from dept_emp as de left join salaries as sa
    on de.emp_no=sa.emp_no
    group by dept_no
) as t1 left join
(
    select de.dept_no,de.emp_no,sa.salary
    from dept_emp as de left join salaries as sa
    on de.emp_no=sa.emp_no
) as t2
on t1.dept_no=t2.dept_no and t1.salary=t2.salary
order by t2.dept_no asc