基于两个表在MySQL中选择

Question

我有两个表。

疾病

----------------------------- 
| ID | NAME    | 
----------------------------- 
| 1 | Disease 1   | 
| 2 | Disease 2   | 
| 3 | Disease 3   | 

diseases_symptoms

----------------------------- 
| DISEASE_ID | SYMPTOM_ID | 
----------------------------- 
| 1   | 1   | 
| 1   | 2   | 
| 1   | 3   | 
| 1   | 4   | 
| 2   | 1   | 
| 2   | 2   | 

我想选择具有症状1或2，3或4。

我已经尝试了所有diseses：

SELECT * 
FROM diseases_symtoms 
WHERE (symptoms = '1' OR symptoms = '2') 
    AND (symptoms = '3' OR symptoms = '4') 

And：

SELECT * 
    FROM diseases_symtoms 
    WHERE symptoms IN ('1','2') 
    AND symptoms IN ('3','4') 

...但它不工作。

Answer 1

请记住，SELECT一次只能检查一行。这两个查询的行为就好像您可以同时检测1和3（例如），这是不可能的。

要同时考虑多个行，您可以加入表中的两个单独的副本，或者尝试一个分组这样的：

SELECT diseases.* 
FROM diseases 
INNER JOIN diseases_symptoms ON (disases_symptoms.disease_id = diseases.disease_id) 
GROUP BY diseases.disease_id 
HAVING SUM(IF(symptoms = 1 OR symptoms = 2, 1, 0) > 0 AND SUM(IF(symptoms = 3 OR symptoms = 4, 1, 0) > 0 

Answer 2

SELECT d.* FROM diseases AS d 
INNER JOIN disease_symptoms AS s1 ON s1.DISEASE_ID = d.ID WHERE SYMPTOM_ID IN (1, 2) 
INNER JOIN disease_symptoms AS s2 ON s2.DISEASE_ID = d.ID WHERE SYMPTOM_ID IN (3, 4) 
GROUP BY d.ID 

Answer 3

你可以试试...

SELECT DISTINCT * 
    FROM diseases 
    WHERE EXISTS (SELECT * 
         FROM disease_symptoms 
         WHERE disease.disease_id = disease_symptoms.disease_id AND 
          symptom_id IN (1,2)) AND 
      EXISTS (SELECT * 
         FROM disease_symptoms 
         WHERE disease.disease_id = disease_symptoms.disease_id AND 
          symptom_id IN (3,4));