从多对多表中选择具有所有这些详细信息的项目……适当的解决方案？

Question

(PHP/MySQL) Let me preface by saying that this works, and that I'm wondering if there's a cleaner way. This seems a bit... brutish? And I'm not known for my efficient coding. :-/

TABLE (many-to-many)
scenid   mapid
  AA      01
  AA      02
  AA      04
  BB      01
  BB      04
  CC      02
  CC      03
  CC      05
  DD      01
  DD      02
  DD      03

You are searching for scenarios that have ALL of these maps: 01, 02, 03
(having all + some others is ok)

My current method is to use this query:

SELECT scenid, COUNT(scenid) FROM `TABLE`
WHERE mapid IN ( 01, 02, 03 )
GROUP BY scenid ORDER BY COUNT(scenid) DESC, scenid ASC

Which would produce this list:

scenid  COUNT
  DD      3
  AA      2
  CC      2
  BB      1

Then, within PHP, I run a while loop that copies scenids to a new array so long as the COUNT is still == the total number of mapids (in this case, 3). This trims off all results that contain only some of the required maps.

Again, this works. But I'm still very new to MySQL & PHP (and they're my first steps into programming) so I'm always wondering if my " works " is another person's " did you see that idiot just reinvent the wheel? lulz "

Is there a better way? Thank you for your time.

Answer 1

Your query looks ok, but you could do the filtering in SQL using HAVING clause:

SELECT scenid, COUNT(*)
FROM `TABLE`
WHERE mapid IN ( 01, 02, 03 )
GROUP BY scenid
HAVING COUNT(*) = 3
ORDER BY scenid ASC