IN for many elements

Question 1

This problem is called Relational Division.

SELECT  a.Name
FROM    Site a
        INNER JOIN SiteWidget b
            ON a.ID = b.Site_ID
        INNER JOIN Widget c
            ON b.Widget_ID = c.ID
WHERE   c.Name IN ('comments','mail')
GROUP   BY a.Name
HAVING  COUNT(*) = 2

SQLFiddle Demo

if uniqueness was not enforce on widget_id for every site_id, DISTINCT keyword is needed.

SELECT  a.Name
FROM    Site a
        INNER JOIN SiteWidget b
            ON a.ID = b.Site_ID
        INNER JOIN Widget c
            ON b.Widget_ID = c.ID
WHERE   c.Name IN ('comments','mail')
GROUP   BY a.Name
HAVING  COUNT(DISTINCT c.Name) = 2

Other Link

SQL of Relational Division

Question 2

Here's one way to do it - use an extra join so that you can look for combinations of 2 widgets:

SELECT * FROM Site s
INNER JOIN SiteWidget w1 ON (s.id = w1.site_id)
INNER JOIN SiteWidget w2 ON (s.id = w2.site_id)
WHERE w1.widget_id=2 and w2.widget_id=4;

Question 3

Try:

SELECT * FROM Site
INNER JOIN SiteWidget SW1
    ON SW1.widget_id = 2 
    AND Site.id = SW1.site_id
INNER JOIN SiteWidget SW2
    ON SW2.widget_id = 4
    AND Site.id = SW2.site_id

Question 4

You can use this if you want filter by widget name

SELECT
    S.id,
    S.name 
FROM Site S
    JOIN SiteWidget SW
        ON S.id = SW.site_id
    JOIN Widget W
        ON SW.widget_id = W.id
WHERE W.name IN ('comments', 'mail')
GROUP BY S.Id,S.name
HAVING COUNT(DISTINCT W.name) = 2

or if you want to filter by widget id

SELECT
    S.id,
    S.name 
FROM Site S
    JOIN SiteWidget SW
        ON S.id = SW.site_id
WHERE SW.widget_id IN (2, 4)
GROUP BY S.Id,S.name
HAVING COUNT(DISTINCT SW.widget_id) = 2

Question 5

Got to join twice

SELECT * FROM Site 
inner JOIN SiteWidget m ON Site.id = m.site_id and m.widget_id = 4
inner Join SiteWidget c ON Site.id = c.site_id and c.widget_id = 2

Question 6

Literally, you would need two different JOINs:

SELECT * FROM Site
    JOIN SiteWidget AS mail     ON (Site.id = mail.site_id AND mail.widget_id = 4)
    JOIN SiteWidget AS comments ON (Site.id = comments.site_id AND comments.widget_id = 2);

If you are sure that the SiteWidget table has no duplicates, e.g. because (site_id, widget_id) is primary key as is usually done for MtM relations, then you can also use HAVING: this is MySQL syntax:

SELECT Site.* FROM Site
    JOIN SiteWidget ON (SiteWidget.site_id = Site.id AND widget_id IN (2,4))
    GROUP BY Site.id HAVING COUNT(*) = 2;

since, due to uniqueness, the only possibility for a site to appear twice is to have both widgets. Some believe this to be an abuse of GROUP BY, and some SQLs (PostgreSQL, if I remember correctly) will require Site's fields to appear in GROUP BY or an aggregate functions in SELECT even if they functionally depend from the group-by column Site.id.

I find the first formula to be clearer and safer, and, I expect, more or less as fast as the second.

This is both because the many-to-many join table is very small (and index-covered to boot), and because this kind of operation was standard from day one, and is one of the most optimized. For example, I expect checks for widget_id 2 and 4 to run in parallel with a single logical read of the SiteWidget table in the join buffer. Even if they didn't, they would likely be loaded in parallel with a single physical read, the other hitting a SQL cache or, at the very least, the IOSS cache.

You might also try this slight variation, which should be the faster:

SELECT Site.* FROM Site
    JOIN SiteWidget AS mail     ON (Site.id = mail.site_id AND mail.widget_id = 4)
    JOIN SiteWidget AS comments ON (mail.site_id = comments.site_id AND comments.widget_id = 2);

which ought to run the main JOIN against the smallest SiteWidget table, and come out with id lookups into Site. This is actually what is likely to get done even if you word the query as in the first instance.

The first formula is perhaps easier to extend by copying and pasting if you ever need to add, say, the polling widget.

Question 7

Here is another way, Fiddle (Thanks @JW. for the fiddle tables & data)

select s.id, s.name
from site s join (
   select sw.site_id, count(w.id) cnt
   from SiteWidget sw join widget w on sw.widget_id = w.id 
   where w.id in (2,4) 
   group by sw.site_id
) T on s.id = T.site_id and T.cnt = 2