mysql select distinct n row that has a certain value

Question

Considering here is my table query:

id    name  number   Code
1     red     1       A
2     red     3       B
3     blue    3       C
4     blue    5       A
5     purple  2       D
6     yellow  3       D
7     yellow  4       C

Now I need to query to get 2 random row such that there is 1 name is red and 1 number is 3, kinda like this:

SELECT * FROM table WHERE name = "red" LIMIT 1 and number = 3 LIMIT 1

So like row 1+3,1+6 or 2 + any other row. Here is my query:

SELECT * FROM table
        group by name,number
        having count(name="red") = 1
             and count(number=3) = 1
ORDER BY RAND()
LIMIT 2;

However, it seems like it just query the row randomly and not satisfying my requirement. Can anyone show me what is wrong? Thank you.

Answer 1

If you can live with odd formatting...

select x.id x_id
 , x.name x_name
 , x.number x_number
 , x.code x_code
 , y.id y_id
 , y.name y_name
 , y.number y_number
 , y.code y_code
    
   from my_table x
   join my_table y
     on y.id <> x.id
  where x.name = 'red'
    and y.number = 3
 order by rand()
 limit 1;

https://www.db-fiddle.com/f/6JmLKq1RwaPrSwS3zx4Qmt/0

Previously, I posted this solution, but it too has some flaws, I think. But TB liked it, so I'll keep it here...

select * 
   from my_table where name = 'red'
  union distinct
select * 
   from my_table where number = 3
 order by rand()
 limit 2

Answer 2

I think that this will do what you want:

select t1.*
from tablename t1 
inner join (
  select t1.id id1, t2.id id2
  from tablename t1 inner join tablename t2
  on t2.id > t1.id 
  and ('red' in (t1.name, t2.name)) + ('3' in (t1.number, t2.number)) = 2
  order by rand() limit 1
) t2 on t1.id in (t2.id1, t2.id2)

Note that the row with the highest probability to be returned is id = 2 , because it can be combined with any other row of the table.
See the demo .