从一个表中选择count（id），其中另一个表中存在该ID

Question

EDIT: Detailed description

Detailed Description:

*in table_a there are 100 members but only 50 of them have records in table_b and only 25 have records in table_b where approved = 1 THEREFORE the value I will need returned by the query is 25*

Hey everyone here is the query I am trying to resolve it will need to return a single result count so I can access with mysql_result($query, 0).

SELECT COUNT(id) FROM table_a WHERE (THIS IS WHERE I AM STUCK)

I need to check if the( count of memberID in table_b WHERE memberID matching each id in table_a and approved in table_b = 1) - is greater than 1

The final result needs to be a count of the number of members that have an entry in table_b.

Sample of table columns that need to access

table_a
-----------------
id

table_b
------------------
id
memberID
approved

Let me know if you need any more details.

Answer 1

I'm assuming you want to get the number of records in table_a that match to table_b, as long as the value in table_b is approved. Therefore, I would join the 2 tables. The inner join ensures that you only consider rows that are in both tables, and the where statement takes care of the approved requirement.

select count(a.id) 
from table_a a
join table_b b
on a.id = b.memberID
where b.approved=1

Answer 2

 SELECT b.ID, a.ID 

 FROM table_a a, table_b b 

 WHERE a.ID = b.ID AND b.ID = 1

Answer 3

I'm a bit rusty in SQL, but can this be achieved by specifying the two tables in the query like this:

SELECT COUNT(DISTINCT table_a.id) 
  FROM table_a, table_b 
  WHERE table_a.id = table_b.MemberID 
    AND table_b.approved = 1;

I believe that meets your select criteria.

Answer 4

SELECT COUNT(*) AS cnt
FROM table_a AS a
WHERE EXISTS
      ( SELECT *
        FROM table_b AS b
        WHERE b.memberID = a.id
          AND b.approved = 1
      )

Answer 5

Problem Solved

Had to think of it backwards

    SELECT COUNT( DISTINCT memberID )
    FROM table_b
    WHERE approved =1

I do not need to even look at table_a seeing as I am counting the memberID based on table_b

Sometimes the solution is so simple and right in front of you.

Thanks for all the help! I hope this helps other in the future.