MySQL从多对多关系中过滤许多元素

Question

I'm currently working on objects that are linked through a many-to-any relationship (it means that many objects can be linked to... not many objects of the same type, but many objects which type can vary).

But, for this question, a many-to-many will be plenty enough.

So, let's say I have a ObjectA table, with the following :

CREATE TABLE (
    id INT(11) NOT NULL, 
    label VARCHAR(30) NOT NULL,
    -- primary keys and all the stuff
);

CREATE TABLE ObjectB (
    id INT(11) NOT NULL,
    label VARCHAR(30) NOT NULL,
    -- primary keys and all the stuff
);

CREATE TABLE ObjectA_ObjectB (
    objectA_id INT(11) NOT NULL,
    objectA_type VARCHAR(250) NOT NULL, -- for the many to any
    objectB_id INT(11) NOT NULL,
    -- primary keys and all the stuff
);

Let's say I want to filter the ObjectA with 2 or three elements from ObjectB. You will think that I should use a IN :

SELECT *
    FROM ObjectA a
        LEFT JOIN ObjectA_ObjectB relation ON a.id = relation.objectA_id AND "ObjectA" = relation.objectA_class
        LEFT JOIN ObjectB b ON relation.objectB_id = b.id
    WHERE b.id IN (1, 2, 3);

But, the thing is, with this request, it gets all the objectAs that are linked with at least ONE of the ObjectBs searched (here, 1, 2, and 3). But, what I want, are those that have all the ObjectB's 1, 2 and 3.

I already tried some things (like the ALL possibility, or pondering to make a first filter outside of SQL), but it didn't give the expected results.

Any ideas ?

---

To sum up, I'd like to be able to do a bit like the issues from GitHub and how they are filtering their label. If you pick up a bunch of labels, only the issues that have all the selected labels are returned, rather than all the issues that have at least one label.

Thanks !

Answer 1

From your comments it sounds like you don't want LEFT JOIN which will return ALL ObjectA s which match the WHERE clause. As for the relation condition you have to count the matching rows:

  SELECT *
    FROM ObjectA a
    JOIN ObjectA_ObjectB relation 
      ON a.id = relation.objectA_id 
     AND "ObjectA" = relation.objectA_class
    JOIN ObjectB b 
      ON relation.objectB_id = b.id
GROUP BY a.id
  HAVING COUNT(CASE WHEN b.id IN (1,2,3) THEN 1 ELSE NULL END)=3;

OR:

  SELECT *
    FROM ObjectA a
    JOIN ObjectA_ObjectB relation 
      ON a.id = relation.objectA_id 
     AND "ObjectA" = relation.objectA_class
    JOIN ObjectB b 
      ON relation.objectB_id = b.id
     AND b.id IN (1,2,3)
GROUP BY a.id
  HAVING COUNT(*)=3;

If you don't actually need the ObjectB data in this query you can simplify this even further:

  SELECT a.*
    FROM ObjectA a
    JOIN ObjectA_ObjectB relation 
      ON a.id = relation.objectA_id 
     AND "ObjectA" = relation.objectA_class
     AND relation.objectB_id IN (1,2,3)
GROUP BY a.id
  HAVING COUNT(*)=3;

This assumes that you can only have one of each ObjectB.id linked to an ObjectA through the pivot table.

Answer 2

A solution is to count the matching conditions in an having clause.

SELECT *
FROM ObjectA a
    JOIN ObjectA_ObjectB relation ON a.id = relation.objectA_id AND "ObjectA" = relation.objectA_class
    JOIN ObjectB b ON relation.objectB_id = b.id
WHERE b.id IN (1, 2, 3) GROUP BY a.id HAVING COUNT(b.id)=3;

See sqlfidle example .