Question

如果我有桌子

`account`, `account_contact`, `contact_event` and `event`

以下FK存在

`account_contact`: `account_id`,`contact_id`
`contact_event`: `contact_id`, `event_id`

如何查询总票数的帐户出席频率＆＃39;链接到帐户和事件的所有联系人，以及该帐户联系人的不同事件数。

我有以下内容但JOIN导致第一个子查询返回具有相同值的重复Ticket's Sold行。

SELECT a1.id,`Tickets Sold`, a2.`Distinct Events Attended` FROM

(SELECT a.id, count(*) `Tickets Sold`
 FROM account a JOIN account_contact ac
     ON ac.active = 1 AND a.id=ac.account_id
 JOIN contact_event ce
     ON ac.contact_id=ce.contact_id
 JOIN event e
     ON ce.event_id=e.id AND ce.role = 'Checked In'
 WHERE e.date_start >= '2015-12-31' AND e.date_end < '2017-01-01' 
 GROUP BY a.id ) a1

LEFT JOIN

(SELECT a.id,count(*) `Distinct Events Attended`
 FROM account a
 JOIN account_contact ac
     ON a.id=ac.account_id AND ac.active = 1
 JOIN contact_event ce
     ON ac.contact_id=ce.contact_id AND ce.role = 'Checked In'
 JOIN event e
     ON ce.event_id=e.id
 WHERE e.date_start >= '2015-12-31' AND e.date_end < '2017-01-01'    
 GROUP BY ac.id ) a2

ON a1.id=a2.id

显示问题的示例输出：正如您所见，Distinct Events正在返回针对该特定事件销售给公司的门票数量，因此在示例中有8个事件已参加，

AccountId   Tickets  DistinctEvents
38110518    15       5
38110518    15       1
38110518    15       3
38110518    15       2 
38110518    15       1
38110518    15       1
38110518    15       1
38110518    15       1

相反，我应该回来：

AccountID   Tickets    DistinctEvents
38110518    15         8

编辑：示例数据

SELECT id,name FROM account;
id  name
100 Big Company


SELECT id,first_name FROM contact;
id  first_name
200 John
800 Other


SELECT contact_id,event_id FROM contact_event;
contact_id  event_id
200 300
200 500
800 300


SELECT account_id,contact_id FROM account_contact;
account_id  contact_id
100 200
100 800


SELECT id,name FROM event;
id   name
300  Big Event
500  Small Event

应该返回：

AccountID   Tickets    DistinctEvents
100         3          2

Answer 1

试试这个：

select
    a.account_id,
    count(event_id) tickets,
    count(distinct event_id) distinct_events
from account_contact a
inner join contact_event e
on a.contact_id = e.contact_id
group by a.account_id;

MySQL按两个不同的维度进行分组

1 个答案: