用于获取列匹配（x AND y AND z）的所有行的查询或代码解决方案

Question

Basically, we know that SQL IN is used to replace many OR conditions (x OR y OR z). But in this case I need a query, or a code solution to select all rows that contain (x AND y AND z). Something like http://www.mobile.de/ is using. On their site, when searching for vehicles the more features you select the less results you get. So they're obviously not using IN in their query. A simplified version of my database would be this.

Adverts table:

--------------
| id | title |
--------------

Features table:

---------------
| id | f_name |
---------------

AdvertsFeatures table:

-------------------------------
| id | advert_id | feature_id |
-------------------------------

For the features search filter I do a LEFT JOIN with the AdvertsFeatures table and apply the following WHERE clause:

WHERE feature_id IN (x, y, z)

Thus, the Adverts that have of the selected features come up. But like I said, I need a query or a code solution that gets only the Adverts that have the selected features. A user can select any number of features from a total of around 40 features. That would be a lot of WHERE clause "AND"s, which I already have many of.

How do I go about doing this? It doesn't have to be SQL, it could be a code solution.

Answer 1

This is an example of a "set-within-sets" query. I like to solve these using aggregation and the having clause. In your case, you can do it this way:

select advert_id
from AdvertsFeatures af
where feature_id in (x, y, z)
group by advert_id
having count(*) = 3;

Note that you need to change the 3 to be the number of items in the in list. This also assumes that there are no duplicates in the AdvertsFeatures data (otherwise, you just use count(distinct featureid) in the having clause).

Answer 2

Gordon's query above can be used as an inline view to obtain the advert_id's. Then, you join these advert_id's with the AdvertFeatures table again to get all the adverts, as below:

SELECT af1.*
FROM AdvertsFeatures af1
INNER JOIN  
    (SELECT advert_id
    FROM AdvertsFeatures
    WHERE feature_id in (x, y, z)
    GROUP BY advert_id
    HAVING count(*) = 3;
) af2
ON af1.advert_id = af2.advert_id;