元素重复至少3次

问题描述

是否有一种方法可以在不使用Count和Group By的情况下确定元素在列中是否至少存在3次?

Is there a way to tell if an element is at least 3 times in a column without using Count and Group By?

元组示例:

{t:{name} | ∃s:{姓名，工资}(雇员w s.wage = 50.000∧t.name = s.name)}

{ t : {name} | ∃ s : {name, wage} ( Employee(s) ∧ s.wage = 50.000 ∧ t.name = s.name ) }

在这里，使用元组的关系演算的局限性很明显.

没有CTE，没有分组依据，没有行工具，没有区别，没有计数，没有视图，没有创建，没有插入，没有更改.这些出色的SQL工具都不是.

No CTE's, no group by, no row tools, no distinct, no count, no views, no create, no insert, no alter. None of those awesome SQL tools.

我不想使用Count和Group By的原因是因为我将使用Tuples将其带到Relational Calculus，这不允许使用那些工具.

The reason why I don't want to use Count and Group By is because I'll take this to Relational Calculus with Tuples, which doesn't allow those tools.

例如:

假设有一个表ORDER(Id_article，Id_Provider)，其中两个ID都是外键.

Suppose there is a table ORDER (Id_article, Id_Provider) where both ID's are foreign keys.

查询:获取至少订购3次的所有文章.

Query: Get all the articles that were ordered at least 3 times.

让表ORDER为:

   Id_Article      Id_Provider

       1               A
       1               B
       1               B
       2               C
       2               C
       3               A

查询结果应仅为元素1，因为它是Id_Article中的3倍.

The result of the query should be only the element 1, as it is 3 times in Id_Article.

推荐答案

可能的解决方案:

让我们致电以下Ids

SELECT O1.id_article AS idFROM ORDER O1JOIN ORDER O2 ON O1.id_article = O2.id_articleMINUSSELECT O3.id_article AS idFROM ORDER O3MINUSSELECT O3.id_article AS idFROM ORDER O3MINUSSELECT O3.id_article AS idFROM ORDER O3MINUSSELECT O3.id_article AS idFROM ORDER O3

SELECT O1.id_article AS idFROM ORDER O1JOIN ORDER O2 ON O1.id_article = O2.id_articleMINUSSELECT O3.id_article AS idFROM ORDER O3MINUSSELECT O3.id_article AS idFROM ORDER O3MINUSSELECT O3.id_article AS idFROM ORDER O3MINUSSELECT O3.id_article AS idFROM ORDER O3

那么我们的关系方程是{ t : {id} | ∃ s : {id} ( Ids(s) ∧ t.id = s.id ) }

Then our relational equation is { t : {id} | ∃ s : {id} ( Ids(s) ∧ t.id = s.id ) }

如果您不能使用MINUS，那么这可能可以帮助

If you can't use MINUS, then maybe this can help Converting aggregate operators from SQL to relational algebra

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