Mysql: Multiple Self Join? - php

Question

I'm building a symptom checker. As the user picks symptoms, we suggest other symptoms from ailments that have the user selected symptoms in common until we can narrow down to a specific ailment. I have this table.
ailment_symptoms
+----+----------+-----------+
|id |ailment_id|symptom_id |
+----+----------+-----------+
|1 | 1 | 1 |
|2 | 1 | 2 |
|3 | 2 | 1 |
|4 | 2 | 3 |
|5 | 3 | 3 |
|6 | 3 | 2 |
|7 | 4 | 1 |
|8 | 4 | 2 |
+----+----------+-----------+
If I want to select ailment_ids of entries that have symptom_id 1 and 2, I use this self join query.
SELECT t1.ailment_id
FROM ailment_symptoms t1
JOIN ailment_symptoms t2 ON t1.ailment_id = t2.ailment_id
WHERE t1.symptom_id = '1'
AND t2.symptom_id = '2'
which will return
+----------+
|ailment_id|
+----------+
| 1 |
| 4 |
+----------+
How do I do it when there are more than two symptom_ids. I want to write a php code that will build for as many symptoms as the users enters. code should look like:
$user_symptoms = array($symptom_id_1, $symptom_id_2, $symptom_id_3); //as many symptom as the user picks
$query = "SELECT t1.ailment_id FROM ailment_symptoms t1";
foreach($user_symptoms AS $symptoms){
//here is where we construct the multiple self join and append it to $query
//please replace this comment with appropriate code
}
$query .= "WHERE ";
//loop through $user_symptoms and append t1.symptom_id = '$user_symptom[0]' AND '....'
Please help me replace the comments with the appropriate code.

Answer 1

You can also do this with aggregation. It might be easier for you to construct the queries like this, because it is easier to handle the additional attributes.
SELECT s.ailment_id
FROM ailment_symptoms s
WHERE s.symptom_id in (1, 2)
GROUP BY s.ailment_id
HAVING COUNT(DISTINCT s.symptom_id) = 2;
You just have to be sure that the "2" matches the number of elements in the list in the where clause. Then it will generalize to any number of symptoms.