Am 16.04.2014 02:36, schrieb Gerald Young:
...
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
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