如何纠正元组排序中的错误？ [URI 2992 最高平均工资部门]

Posted 2023-02-16

【中文标题】如何纠正元组排序中的错误？ [URI 2992 最高平均工资部门]

【英文标题】：How can I correct the error in tuple ordering? [URI 2992 Highest Avarage Salary Divisions]

【发布时间】：2021-02-28 05:38:28

【问题描述】：

谁能帮我理解错误？我无法正确排序。错误 25% URI。我的查询（GMB）：

select name_dep, name_div, max(avg_salary)
from(
      select departamento.nome as name_dep, divisao.nome as name_div, COALESCE(ROUND(AVG(vencimento.valor), 2) , 0) as avg_salary
      from departamento, divisao, empregado, vencimento, emp_venc
      where departamento.cod_dep=divisao.cod_dep and empregado.lotacao_div=divisao.cod_divisao and empregado.matr=emp_venc.matr and vencimento.cod_venc=emp_venc.cod_venc
      group by departamento.nome, divisao.nome
      ) as foo
group by name_dep, name_div, avg_salary
order by avg_salary desc;

SQL：https://pastebin.com/7PyF9bDT

URI：https://www.urionlinejudge.com.br/judge/en/problems/view/2992

【参考方案1】：

您希望每个部门的薪水最高的部门。我会推荐distinct on：

select distinct on (de.nome)
    de.nome as name_dep, 
    di.nome as name_div, 
    coalesce(round(avg(ve.valor), 2) , 0) as avg_salary
from departamento de
inner join divisao di    on di.cod_dep = de.cod_dep
inner join empregado em  on em.lotacao_div = di.cod_divisao
inner join emp_venc ev   on ev.matr = em.matr
inner join vencimento ve on ve.cod_venc = ev.cod_venc
group by de.nome, di.nome
order by de.nome, avg(ve.valor) desc

请注意，这使用标准的显式连接，而不是老式的隐式连接； （字面上）几十年前的这种语法不应该在新代码中使用。我还使用了表别名，这使查询更短且更具可读性。

如果你想允许顶部关系，那么另一种方法是使用窗口函数：

select *
from (
    select 
        de.nome as name_dep, 
        di.nome as name_div, 
        coalesce(round(avg(ve.valor), 2) , 0) as avg_salary,
        rank() over(partition by de.nome order by avg(ve.valor) desc) rn
    from departamento de
    inner join divisao di    on di.cod_dep = de.cod_dep
    inner join empregado em  on em.lotacao_div = di.cod_divisao
    inner join emp_venc ev   on ev.matr = em.matr
    inner join vencimento ve on ve.cod_venc = ev.cod_venc
    group by de.nome, di.nome
) t
where rn = 1
order by name_dep