查询以获取特定区域内有多少客户的百分比

Question

我有一个x，y coord客户表，我需要查找某个区域内有多少客户居住的百分比（50,50）

查询测试数据少于50的任何人：

SELECT X_Coord, Y_Coord
from customerlocation
count where (X_Coord < 50 or Y_Coord < 50)

50岁的人：

SELECT X_Coord, Y_Coord
from customerlocation
count where (X_Coord = 50 or Y_Coord = 50)

如何对这些查询进行分组，以获得一定比例（50,50）生活在（50,50）范围内的人？

Answer 1

为每项要求单独计算并除以总计数

select (a.count / count(*)) * 100, (b.count / count(*)) * 100
from 
(select count(*) as count
  from customerlocation where x_coord < 50 or y_coord < 50) as a,
(select count(*) as count
  from customerlocation where x_coord = 50 or y_coord = 50) as b,
customerlocation

Answer 2

的MySQL

计算X，Y坐标值：

Strict markdown style by default

Answer 3

试试这个。

select
cast(sum(A.lessthan50) as float)/cast((sum(A.lessthan50) + 
sum(A.anyoneat50)) as float)*100 lessthan50, 
cast(sum(A.anyoneat50) as float)/cast((sum(A.lessthan50) + 
sum(A.anyoneat50)) as float)*100 anyoneat50
from
(
select 
case when (x < 50  or y < 50) then 1 else 0 end as lessthan50,
case when (x = 50 or y = 50) then 1 else 0 end as anyoneat50
from cordinates
) A

Answer 4

我认为这是最简单的方法：

select avg( (X_Coord = 50 or Y_Coord = 50) )
from customerlocation
where X_Coord <= 50 or Y_Coord <= 50;

这看起来有点棘手，所以这里有一个稍微冗长的替代方案：

select sum( (X_Coord = 50 or Y_Coord = 50) ) / sum( (X_Coord < 50 or Y_Coord < 50) )
from customerlocation
where X_Coord <= 50 or Y_Coord <= 50;