基于两个表在MySQL中选择

Question

我有两张桌子。

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           |

我想选择症状为1或2和3或4的所有疾病。

我试过了：

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

和：

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

......但它不起作用。

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你可以同时检测到1和3 （例如），这是不可能的。

要一次考虑多个行，您可以加入表的两个单独副本，或尝试这样的分组：

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));