Select count(*), where 2 rows with different values exists I have following table structure and data:

+-----------+-------+
|type       | value |
+-----------+-------+
|first_name | john  |
+-----------+-------+
|last_name  | doe   |
+-----------+-------+
|last_name  | smith |
+-----------+-------+


I need to get count (in my case it's 1), if table contain data John Doe. Query like:

SELECT COUNT(*) FROM tablename WHERE (type='first_name' AND value='john') AND (type='last_name' AND value='doe')

returns zero count. Don't understand, how to create this combined query.
I think I found the solution for this case. My solution is to use SUM() aggregation with IF() condition. The final logic is substitute COUNT(*) to boolean - have rows or empty query. This query will return results only if it match both conditions:

SELECT SUM(IF((value='john' AND type='first_name') OR (value='doe' AND type='last_name'), 1, 0)) AS cnt FROM tablename HAVING cnt=2;


If I want to select all data where first name is John and last name is Doe OR last name is Smith, query will be:

SELECT sum(IF((value='john' AND type='first_name') OR (value='doe' AND type='last_name'), 1, 0)) AS first_last, SUM(IF(value='smith' AND type='last_name', 1, 0)) AS last FROM tablename HAVING first_last=2 OR last=1;