如何获取多个表之间一对多关系中的实例数？

Question

I have a query where I am attempting to find multiple instances of a custtype for a specific date. The following query works fine as is:

select idm, count(*) totals
from customer.purch a
where exists (select null
          from customer.name b
          where b.idad = a.id
          and custdate = to_date('5/11/2017', 'MM/DD/YYYY')
          and custtype = 'R'
          )
group by idm
having count(*) > 1;

The issue I am having however, is I am not sure how to expand the above query to find different combinations of custtype on the given date. For example, how can i get the query to give me the idm of a record where at least one custype is R and at least one other custype is 'Z'? I figure a self join on the customer.name field might be the correct way to go, but I am not sure how to implement it. Thank you for your time.

Sample Data

Table customer.purch a
_______________
|idm  | id     |
| 1   | 896    |
| 2   | 207    |
| 3   | 359    |
________________

Table customer.name b
______________________________
|idad  |  custdate   |custtype|
| 896  |  5/11/2017  |   R    |
| 896  |  5/11/2017  |   Z    |
| 207  |  5/11/2017  |   R    |
| 207  |  5/11/2017  |   X    |
| 359  |  5/11/2017  |   R    |
| 359  |  5/11/2017  |   Z    |
| 359  |  5/11/2017  |   R    |
| 359  |  5/11/2017  |   R    |
_______________________________


Output 
______________
|IDM  | count |
| 1   |  2    |
| 3   |  4    |
_______________

Answer 1

If I understand correctly, you can use aggregation and a having clause:

select p.idm, count(*) totals
from customer.purch p join
     customer.name n
     on p.id = n.idad
where n.custdate = date '2017-05-11'
group by p.idm
having sum(case when n.custtype = 'R' then 1 else 0 end) > 0 and
       sum(case when n.custtype = 'Z' then 1 else 0 end) > 0;