[英]Select in MySQL based on two tables
我有兩張桌子。
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 |
我想選擇症狀為1或2和3或4的所有疾病。
我試過了:
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。
請記住,SELECT一次只能檢查一行。 這兩個查詢的行為就好像你可以同時檢測到1
和3
(例如),這是不可能的。
要一次考慮多個行,您可以加入表的兩個單獨副本,或嘗試這樣的分組:
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
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
你可以試試......
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));
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