Mysql: 多个自连接？

Question

我正在构建一个症状检查器。当用户选择症状时，我们会建议具有用户 select 共同症状的疾病的其他症状，直到我们可以缩小到特定疾病。我有这个 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        |
+----+----------+-----------+

如果我想要 select ailment_ids 个具有 symptom_id 1 和 2 的条目，我使用这个自连接查询。

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'

这将 return

+----------+
|ailment_id|
+----------+
| 1        |
| 4        |
+----------+

超过两个symptom_id怎么办。我想编写一个 php 代码，该代码将针对用户输入的尽可能多的症状进行构建。代码应如下所示：

$user_symptoms = array($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 '....'

请帮我用合适的代码替换注释。

Answer 1

您也可以通过聚合来做到这一点。您可能更容易构造这样的查询，因为它更容易处理附加属性。

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;

您只需确保“2”与 where 子句中列表中的元素数相匹配。然后它将概括为任意数量的症状。

Mysql: 多个自连接？

Mysql: Multiple Self Join?

php

mysql

self-join