计算有多少客户有 1 项服务，有多少客户有 2 项服务

Question

我有这张桌子

id_cliente | serv_cuidpess | serv_ali | serv_hab | serv_roupa

125       | 1              | 2        | 1        | 1  
126       | 2              | 1        | 2        | 1

这意味着 id_cliente = customerID 和其他 serv_ = services 其中 1 = 客户拥有该服务，2 = 他没有。

我想要实现的是计算有多少客户有 1 项服务，有多少客户有 2 项服务，有多少客户有 3 项服务等等。我正在使用 PHP / MySQL

Answer 1

不确定您到底在寻找什么，有 2 个案例

当每个客户 = 1 时，获取所使用的总服务。

当它 = 1 时获取每项服务的客户数

通过以上内容，您可以尝试以下查询，但最好在问题中提供有关预期输出的更多信息。

对于第一种情况，您可以这样做

select
s.id_cliente,
sum(s1.ser1+s1.ser2+s1.ser3+s1.ser4) as tot_services
from services s
inner join
(
  select
  id_cliente,
  case  when serv_cuidpess = 1 then 1 else 0 end as ser1,
  case  when serv_ali = 1 then 1 else 0 end as ser2,
  case  when serv_hab = 1 then 1 else 0 end as ser3,
  case  when serv_roupa = 1 then 1 else 0 end  as ser4
  from services 


)s1 on s1.id_cliente = s.id_cliente
group by s.id_cliente;

对于第二种情况，您可以这样做

select
serv1_client_count,
serv2_client_count,
serv3_client_count,
serv4_client_count
from services s
inner join
(
  select 
  s1.id_cliente,
  sum(s2.ser1) as serv1_client_count,
  sum(s2.ser2) as serv2_client_count,
  sum(s2.ser3) as serv3_client_count,
  sum(s2.ser4) as serv4_client_count
  from services s1
  inner join
  (
    select
    id_cliente,
    case  when serv_cuidpess = 1 then 1 else 0 end as ser1,
    case  when serv_ali = 1 then 1 else 0 end as ser2,
    case  when serv_hab = 1 then 1 else 0 end as ser3,
    case  when serv_roupa = 1 then 1 else 0 end  as ser4
    from services 
  )s2 on  s2.id_cliente = s1.id_cliente
)t on t.id_cliente = s.id_cliente

DEMO

更新

来自评论

输出应该显示有多少有 1 服务而不是服务 1.. 如何 许多人有 2 项服务而不是 2 项服务

你可以这样做

select
sum(client_count) as client_count,
serv_count
from services s
inner join
(
  select 
  s1.id_cliente,
  1 as client_count,
  sum(s2.ser1+s2.ser2+s2.ser3+s2.ser4) as serv_count
  from services s1
  inner join
  (
    select
    id_cliente,
    case  when serv_cuidpess = 1 then 1 else 0 end as ser1,
    case  when serv_ali = 1 then 1 else 0 end as ser2,
    case  when serv_hab = 1 then 1 else 0 end as ser3,
    case  when serv_roupa = 1 then 1 else 0 end  as ser4
    from services 
  )s2 on  s2.id_cliente = s1.id_cliente
  group by s1.id_cliente
)t on t.id_cliente = s.id_cliente
group by serv_count

DEMO

Answer 2

既然你提到了 serv_ 作为进程，我假设所有这些(serv_cuidpess + serv_ali + serv_hab + serv_roupa ) 形成进程。

要获得总和，您可以使用+ 运算符直接添加列HAVINGclause 用于过滤虚拟列，即 TotalProcess

试试下面的查询

 select COUNT(id_cliente) AS NoOfUsers,(serv_cuidpess + serv_ali + serv_hab + serv_roupa ) as TotalProcess from PROCESSTABLE GROUP BY TotalProcess HAVING TotalProcess > 0;