Question

我正在开发一个PHP + MySQL项目，需要知道如何进行SQL查询。下面是一个表格草图：

http://s24.postimg.org/na86w4do4/a_1.jpg

ms_diseases表存储疾病名称。 ms_characteristics存储可能来自某种疾病的特征。 ms_diseases_characteristics表跨越疾病与特征之间的关系

这是一个例子：

http://s21.postimg.org/sg1f7kfti/b_1.jpg

“偏头痛”疾病（ms_diseases）在ms_diseases_characteristics表中具有1和2个特征。

现在的问题是：如何进行查询，使用值1和2仅返回具有1和2特征的疾病？那具体案例是“偏头痛”？

示例：

Characteristics – 1, 2 – return -> MIGRAINE
Characteristics – 1 – return > NOTHING
Characteristics – 1, 4, 5 – return > FLU
Characteristics – 1, 4 – return > NOTHING

如何进行返回具有此类特征的疾病的查询？

Answer 1

SELECT disease_name
FROM
(
    SELECT  a.disease_name, c.totalCount
    FROM    ms_diseases a
            INNER JOIN ms_diseases_characteristics b
                ON  a.ID = b.disease_ID 
            INNER JOIN 
            (
                SELECT  disease_ID, COUNT(*) totalCount
                FROM    ms_diseases_characteristics
                GROUP   BY disease_ID
            ) c ON  b.disease_ID  = c.disease_ID 
    WHERE   b.characteristic_id IN (1,4,5)         -- <<== list of charateristics
    GROUP   BY a.disease_name                                       ^^
    HAVING  COUNT(*) = c.totalCount AND                             ^^
            COUNT(DISTINCT b.characteristic_id) = 3 -- << # of parameters
) s

SQLFiddle Demo

Answer 2

SELECT CASE WHEN A.COUNT = B.FOUND_DISES THEN M.DISEASE_NAME ELSE 'NOT FOUND' END AS     DISEASE
FROM
(
SELECT COUNT(*) AS COUNT, DISEASE_ID
FROM MS_DISEASES_CHARACTERISTICS
GROUP BY DISEASE_ID
) A
JOIN
(
SELECT COUNT(*) AS FOUND_DISES, DISEASE_ID
FROM MS_DISEASES_CHARACTERISTICS
WHERE CHARACTERISTIC_ID IN (<VALUE YOU PASS>)
GROUP BY DISEASE_ID
) B
ON WHERE A.DISEASE_ID = B.DISEASE_ID
JOIN MS_DISEASES AS M
ON D.ID = A.DISEASE_ID

Answer 3

SELECT disease_name
FROM ms_diseases
WHERE id = (SELECT disease_id FROM (
  SELECT disease_id, count(disease_id) CT
  FROM 
      ms_diseases_characteristics AS dc,
      ms_characteristics AS c 
  WHERE
      c.id = dc.characteristic_id AND
      c.id IN (1, 2)
  GROUP BY disease_id
  ORDER BY CT DESC) temp LIMIT 1);

或者如果您想使用特征名称：

SELECT disease_name
FROM ms_diseases
WHERE id = (SELECT disease_id FROM (
  SELECT disease_id, count(disease_id) CT
  FROM 
      ms_diseases_characteristics AS dc,
      ms_characteristics AS c 
  WHERE
      c.id = dc.characteristic_id AND
      c.characteristic_name IN ('Headache','Frequent pain')
  GROUP BY disease_id
  ORDER BY CT DESC) temp LIMIT 1);

行和表之间的关系 - mySQL

3 个答案: