MySQL,PHP: count results per condition + get the total result that satisfies the conditions

Question

querytest table:

1. xxx | yyy
2. 999 | 888
3. 999 | 787
4. 111 | 222

query:

SELECT *,
       SUM(CASE WHEN XXX LIKE '999' AND YYY RLIKE '[0-9]8[0-9]' THEN 1 ELSE 0 END) AS Count999,
       SUM(CASE WHEN XXX LIKE '111' AND YYY RLIKE '[0-9]2[0-9]' THEN 1 ELSE 0 END) AS Count111
    FROM querytest
    WHERE 
    ( XXX LIKE '999' AND YYY RLIKE '[0-9]8[0-9]') 
    OR 
    ( XXX LIKE '111' AND YYY RLIKE '[0-9]2[0-9]')

result:

Array
(
    [0] => 999
    [xxx] => 999
    [1] => 888
    [yyy] => 888
    [2] => 2
    [Count999] => 2
    [3] => 1
    [Count111] => 1
)

clarification:

I'm using php.and i want a query that

• let me know how many results I can get from ( XXX LIKE '999' AND YYY RLIKE '[0-9]88[0-9]') and how many results i can get from ( XXX LIKE '111' AND YYY RLIKE '[0-9]23[0-9]') and so on. like in the previous query.

• gets me all the results like if im using this query:

 SELECT * FROM querytest WHERE ( XXX LIKE '999' AND YYY RLIKE '[0-9]8[0-9]') OR ( XXX LIKE '111' AND YYY RLIKE '[0-9]2[0-9]') 

my question

how can I combine the two previous points into one query? plus the ability of using LIMIT and ORDER with the second point.

Answer 1

您或者需要进行单独的查询，或者使用UNION。