错误使用内部联接功能/组功能？

Question

I have the following problem with my query:

I have two tables:

1. Customer
2. Subscriber

linked together by customer.id=subscriber.customer_id

in the subscriber table, I have records with id_customer=0 (these are email records, that do not have a full customer account)

Now i want to show how many customers I have per day, and how many subscribers with id_customer, and how many subscribers WITH id_customer=0 (emailonlies i call them)

Somehow, i cannot manage to get those emailonlies. Perhaps it has something to do with not using the right join type. When i use left join, i get the right amount of customers, but not the right amount of emailonlies. When I use inner join i get the wrong amount of customers. Am i using the group function correctly? i think it has something to do with that.

THIS IS MY QUERY:

   `   SELECT DATE(c.date_register),
COUNT(DISTINCT c.id) AS newcustomers,
COUNT(DISTINCT s.customer_id) AS newsubscribedcustomers,
COUNT(DISTINCT s.subscriber_id AND s.customer_id=0) AS emailonlies
FROM customer c
LEFT JOIN subscriber s ON s.customer_id=c.id
GROUP BY DATE(c.date_register)
ORDER BY DATE(c.date_register) DESC
LIMIT 10
;`

Answer 1

I'm not entirely sure, but I think in DISTINCT s.subscriber_id AND s.customer_id=0 , it runs the AND before the DISTINCT , so the DISTINCT only ever sees true and false.

Why don't you just take

COUNT(DISTINCT s.subscriber_id) - (COUNT(DISTINCT s.customer_id) - 1)?

(The -1 is there because DISTINCT s.customer_id will count 0.)

Answer 2

Got it, only risk is that i get no email onlies if there are no customers on this day, becuase of the left join. But this one works:

SELECT customers.regdatum,customers.customersqty,subscribers.emailonlies

FROM ( (SELECT DATE(c.date_register) AS regdatum,COUNT(DISTINCT c.id) AS customersqty FROM customer c GROUP BY DATE(c.date_register) ) AS customers

LEFT JOIN

(SELECT DATE(s.added) AS voegdatum,COUNT(DISTINCT s.subscriber_id) AS emailonlies
FROM subscriber s
WHERE s.customer_id=0
GROUP BY DATE(s.added)
) AS subscribers

ON customers.regdatum=subscribers.voegdatum ) ORDER BY customers.regdatum DESC ;