mysql: counting number of tickets which are open per day basis
You can use conditional sum for this calculation something as

select 
date_format(created,'%d-%m') as created_date , 
sum( case when status='open' then 1 else 0 end )+
sum( case when status='new' then 1 else 0 end )+
sum( case when status='resolved' and date(resolved) &gt; date(created) then 1 else 0 end ) as open
from table_name
where
YEAR(Created) = '2015'
group by created_date ;


Here is a test case

mysql&gt; select * from test ;
+------+---------------------+---------------------+----------+
| id   | created             | resolved            | status   |
+------+---------------------+---------------------+----------+
|    1 | 2015-05-10 00:00:00 | 1970-01-01 00:00:00 | open     |
|    2 | 2015-05-10 00:00:00 | 1970-01-01 00:00:00 | new      |
|    3 | 2015-05-10 00:00:00 | 2015-05-12 00:00:00 | resolved |
|    4 | 2015-05-11 00:00:00 | 1970-01-01 00:00:00 | open     |
|    5 | 2015-05-11 00:00:00 | 1970-01-01 00:00:00 | new      |
|    6 | 2015-05-11 00:00:00 | 2015-05-11 00:00:00 | resolved |
+------+---------------------+---------------------+----------+
6 rows in set (0.00 sec)

mysql&gt; select 
    -&gt; date_format(created,'%d-%m') as created_date , 
    -&gt; sum( case when status='open' then 1 else 0 end )+
    -&gt; sum( case when status='new' then 1 else 0 end )+
    -&gt; sum( case when status='resolved' and date(resolved) &gt; date(created) then 1 else 0 end ) as open
    -&gt; from test
    -&gt; where
    -&gt; YEAR(Created) = '2015'
    -&gt; group by created_date ;
+--------------+------+
| created_date | open |
+--------------+------+
| 10-05        |    3 |
| 11-05        |    2 |
+--------------+------+
2 rows in set (0.00 sec)


UPDATE
The count of open is a incremental as per the comment. So the following should do the trick

select
created_date,
@open:= @open+open_ticket-resolved as open
from(
 select 
 date_format(created,'%d-%m') as created_date , 
 sum( case when status='open' then 1 else 0 end )+
 sum( case when status='new' then 1 else 0 end )+
 sum( case when status='resolved' and ( date(resolved) &gt; date(created) or date(resolved) = date(created)) then 1 else 0 end ) as open_ticket,
 sum(case when status = 'resolved' and date(resolved) = date(created) then 1 else 0 end) as resolved
 from test
 where
 YEAR(Created) = '2015'
 group by created_date
)x,(select @open:=0)r order by created_date ; 


From the above sample data it will have

 +--------------+------+
| created_date | open |
+--------------+------+
| 10-05        |    3 |
| 11-05        |    5 |
+--------------+------+


UPDATE:

Count the previous tickets which got resolved on a given date and deduct that from the current day open ticket, for that you need a left join something as

select
created_date,
@open:= @open+open_ticket-resolved_ticket as open
from(
 select 
 date_format(t.created,'%d-%m') as created_date , 
 sum( case when t.status='open' then 1 else 0 end )+
 sum( case when t.status='new' then 1 else 0 end )+
 sum( case when t.status='resolved' and ( date(t.resolved) &gt; date(created) or date(t.resolved) = date(created)) then 1 else 0 end ) as open_ticket,
 coalesce(tot,0) as resolved_ticket
 from test t left join(
    select count(*) as tot, date(resolved) as resolved from test where status='resolved' group by date(resolved)
 )x on date(x.resolved) = date(t.created) 
 where
 YEAR(t.created) = '2015'
 group by created_date
)x,(select @open:=0)r order by created_date ; 


OUTPUT :

+--------------+------+
| created_date | open |
+--------------+------+
| 10-05        |    3 |
| 11-05        |    5 |
| 12-05        |    7 |
+--------------+------+