mysql - Count all records that does not exist to other table - SQL Query -

i have two(2) tables , i'm trying count records table1 , table1_delta pagename table1_delta not yet listed table1. incase pagename table1_delta listed table1, status must 1 included in count result.

sample table structure:

table1

+-----------+--------+ | pagename  | status | +-----------+--------+ | pagename1 |   2    | | pagename2 |   1    | +-----------+--------+ 

table1_delta

+-----------+ | pagename  | +-----------+ | pagename1 | | pagename2 | | pagename3 | | pagename4 | +-----------+ 

the table sample should return "3".

pagename3 , pagename4 not listed in table1(that returns 2) , pagename2 table1 has status = 1(that returns 1). in total there 3 pagenames table1_delta not listed in table1 , record table1 status = 1. i'm wondering on how query of this? i'm using mysql v5.6.17. thanks!

here alternative solution using joins:

select count(*) table1_delta t1 left join table1 t2 on t1.pagename = t2.pagename t2.status null or t2.status = 1 

here temporary table above query looks like:

+-----------+--------+ | pagename  | status | +-----------+--------+ | pagename1 |  2     |    # row not counted | pagename2 |  1     |    # +1 row has status = 1 , counted | pagename3 |  null  |    # +1 row has status = null , counted | pagename4 |  null  |    # +1 row null , counted +-----------+--------+ 

check out link below running demo.

sqlfiddle