计算表上的实例(Count instances on a table)

计算表上的实例(Count instances on a table)

如果我有一个包含重复实例的表，如果我没有计数功能，我怎么算？

我只有选择，项目，联合，差异，产品，交叉，njoin。 我正在使用WinRDBI 。

表看起来像这样：

孩子

ID| AME| A | 'alice' A | 'jon' A | 'alex' B | 'joe' B | 'mary' C | 'amy'

亲

ID| AME| A | 'Smith' B | 'Johnson' C | 'Meyer'

我想知道父母有两个孩子的情况。

If I have a table with duplicate instances how can I count if I don't have a count function?

All I have is select, project, union, difference, product, intersect, njoin. I am using WinRDBI.

Table looks like this:

Children

ID| AME| A | 'alice' A | 'jon' A | 'alex' B | 'joe' B | 'mary' C | 'amy'

Parent

ID| AME| A | 'Smith' B | 'Johnson' C | 'Meyer'

I want to know how what parent has two children.

最满意答案

当n> 0时，使用差值运算符和(n*n)-n = n仅对于n = 2为真。

对于每个父母，通过重命名儿童的第二个副本[称之为“C1”]，创建他们的孩子的交叉产品[称之为“C”]。 我们称之为“CxC1”

如果（（（从CxC1选择C的属性） - C）= C）那么父母正好有2个孩子[1]

[1]假设参照完整性使父母不能有零孩子。

Use the difference operator and the fact that (n*n)-n = n is true only for n = 2 when n > 0.

For each parent, create a cross product of their Children [call this "C"] with itself by renaming second copy of Children [call this "C1"]. Let's call this "CxC1"

If (((select attributes of C from CxC1) - C) = C) then the parent has exactly 2 children [1]

[1] Assuming referential integrity such that a parent cannot have zero children.