我正在建立症状检查器。当用户选择症状时,我们建议从疾病中选择其他症状,使用户选择共同的症状,直到我们可以缩小到特定疾病为止。我有这张桌子。
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 |
+----+----------+-----------+
如果要选择具有symptom_id 1和symptom_id 2的条目的ailment_id,请使用此自连接查询。
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'
这将返回
+----------+
|ailment_id|
+----------+
| 1 |
| 4 |
+----------+
当有两个以上的symptom_id时,我该怎么办。我想编写一个php代码,该代码将针对用户输入的症状进行构建。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 '....'
请帮助我用适当的代码替换注释。
您也可以通过聚合来做到这一点。这样构造查询可能会更容易,因为处理其他属性会更容易。
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;
您只需要确保“ 2”与where
子句中列表中元素的数量匹配即可。然后它将泛化为任何数量的症状。
本文收集自互联网,转载请注明来源。
如有侵权,请联系[email protected] 删除。
我来说两句