Am 16.04.2014 02:36, schrieb Gerald Young:
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select q.name http://q.name queue, count(t.id http://t.id) number from ticket t left join queue q on q.id http://q.id=t.queue_id left join ticket_state ts on ts.id http://ts.id = t.ticket_state_id group by q.name http://q.name
number of tickets locked and unlocked in each queue:
select q.name http://q.name queue, sum(if(t.ticket_lock_id =1, 1, 0)) unlocked, sum(if(t.ticket_lock_id = 2, 1, 0)) locked from ticket t left join queue q on q.id http://q.id=t.queue_id left join ticket_state ts on ts.id http://ts.id = t.ticket_state_id group by q.name http://q.name
(yeah, you're not going to find *that* one easily in a list).
Thank you Gerald for sharing! (this is also ment for many other answers you gave in this List) regards Fritz