仅当所有连接条件都匹配时，Oracle才会返回行

Question

http://sqlfiddle.com/#!4/bab93d

See the SQL Fiddle example... I have Customers, Tags, and a mapping table. I am trying to implement customer search by tags, and it has to be an AND search. The query is passed a list of tag identifiers (any number), and it has to return only customers that have ALL the tags.

In the example I have used an IN operator, but that corresponds to an OR search and doesn't solve my problem. What should the query look like to be an AND search?

select
  *
from
  customer c
  inner join customer_tag ct on ct.customer_id = c.customer_id
where
  ct.tag_id in (1, 2);

This returns both customers, but only the first customer is tagged with tag 1 and 2.

Answer 1

You could use correlated subquery to get list of all customers:

SELECT  *
FROM  customer c
WHERE c.customer_ID IN
(
     SELECT customer_id
     FROM customer_tag ct
     WHERE ct.customer_id = c.customer_id
       AND ct.tag_id IN (1,2)
     GROUP BY customer_id
     HAVING COUNT(DISTINCT tag_id) = 2
);  

LiveDemo

It is easy to extend just:

WHERE ct_tag IN (1,2,3)
...
HAVING COUNT(DISTINCT tag_id) = 3

Answer 2

JOIN version:

SELECT c.*
FROM  customer c
JOIN (SELECT customer_id
      FROM customer_tag
      WHERE tag_id IN (1,2)
      GROUP BY customer_id
      HAVING MAX(tag_id) <> MIN(tag_id)) ct ON c.customer_id = ct.customer_id

If you have more than 2 different values, use COUNT DISTINCT instead, like this:

SELECT c.*
FROM  customer c
JOIN (SELECT customer_id
      FROM customer_tag
      WHERE tag_id IN (1,2,3)
      GROUP BY customer_id
      HAVING COUNT(DISTINCT tag_id) = 3) ct ON c.customer_id = ct.customer_id

Answer 3

You could do something like

SELECT *
FROM customer c
WHERE 
  2 = (SELECT COUNT(DISTINCT ct.tag_id) 
       FROM customer_tag ct 
       WHERE ct.customer_id = c.customer_id AND ct.tag_id IN (1, 2));

The 2 = ... part should be adjusted according to the number of tags you are trying to search for.

---

Alternative solution - correlated subqueries can have poor performance, so you could join in a table like this

SELECT c.*
FROM customer c
  INNER JOIN (
    SELECT ct.customer_id
    FROM customer_tag ct
    WHERE ct.tag_id IN (1, 2)
    GROUP BY ct.customer_id
    HAVING COUNT(DISTINCT ct.tag_id) = 2
  ) ct ON ct.customer_id = c.customer_id;

Answer 4

Extending the other answers, I will add a version where you don't need to count the tags nor list them twice:

WITH tgs as(
   select distinct tag_id
   from Tags
   where tag_id in (1, 2, 3, 4) --modify only here
) 
SELECT c.*
FROM  customer c
JOIN (SELECT customer_id
      FROM customer_tag
      WHERE tag_id in (select tag_id from tgs)
      GROUP BY customer_id
      HAVING COUNT(DISTINCT tag_id) = (select count(*) from tgs)
     ) ct ON c.customer_id = ct.customer_id

If the tables are rightly normalized(and have pks, etc) you can remove the distinct keywords.