从另一个表添加列，但不影响计数（）

Question

我已经其输出类似

reservation_id | customer_id | name | num_of_ordered_services 
    1   | 1909091202 | John |   2 
    2   | 2512541508 | Jane |   3 

我想查询有多个连接，保留

SELECT reservation.reservation_id, customer.customer_id, customer.name, count(ordered_services.reservation_id) AS num_of_ordered_services 
FROM reservations 
JOIN customers ON reservations.customer_id = customer.customer_id 
LEFT JOIN ordered_services ON reservations.reservation_id = ordered_services.reservation_id 
GROUP BY reservation.reservation_id, customer.customer_id, customer.name 
ORDER BY reservation.reservation_id 

的简单列表添加另一列有关付款的信息，但简单的JOIN，LEFT JOIN会干扰现有的count（）列。像

SELECT reservation.reservation_id, count(payments.reservation_id) AS num_of_payments 
FROM reservations 
LEFT JOIN payments ON reservations.reservation_id = payments.reservation_id 
GROUP BY reservation.reservation_id 
ORDER BY reservation.reservation_id 

reservation_id | num_of_payments 
    1   |  0 
    2   |  2 

但在一个单一的结果。如何实现这一目标？

PS：num_of_payments不是必需的，我只需要知道是否存在某个预留的支付（1,0）。 谢谢

---

TBL结构，没有什么特别的： 保留

reservation_id | customer_id |  added 
    1   | 1909091202 | 2011-11-04 02:37:28 
    2   | 2512541508 | 2011-11-04 14:27:01 

客户

customer_id | name | personal information columns ... 
1909091202 | John |  |  | 
2512541508 | Jane |  |  | 
    ...  | ... |  |  | 

支付

payment_id | reservation_id | customer_id | total |  added 
    1  |  2  | 1909091202 | 199 | 2011-11-04 02:37:28 
    2  |  2  | 2512541508 | 50 | 2011-11-04 14:27:01 

Answer 1

您可以使用子选项作为附加字段。

SELECT reservation.reservation_id, customer.customer_id, customer.name, 
count(ordered_services.reservation_id) AS num_of_ordered_services, 
(SELECT count(*) FROM payments WHERE reservation.reservation_id=payments.reservation_id) AS num_of_payments 
FROM reservations 
JOIN customers ON reservations.customer_id = customer.customer_id 
LEFT JOIN ordered_services ON reservations.reservation_id = ordered_services.reservation_id 
GROUP BY reservation.reservation_id, customer.customer_id, customer.name 
ORDER BY reservation.reservation_id 

Answer 2

Somethi ng如下应该工作：

select 
reservation.reservation_id, 
(case when exists (select * from payments p1 where p1.reservation_id = reservation.reservation_id) then 1 else 0 end) as one_or_many_payments_made 
from reservation 
GROUP BY reservation.reservation_id 
ORDER BY reservation.reservation_id 

但是没有你的数据，这里有一些猜测。