如何创建一个字段不等于 X 的连接？

Question

我有以下数据库表，其中包含有关人员、疾病和药物的信息：

PERSON_T              DISEASE_T               DRUG_T
=========             ==========              ========
PERSON_ID             DISEASE_ID              DRUG_ID
GENDER                PERSON_ID               PERSON_ID
NAME                  DISEASE_START_DATE      DRUG_START_DATE
                      DISEASE_END_DATE        DRUG_END_DATE

我想编写一个查询来查找所有患有 disease_id 52 但未服用药物 34 的人。我该怎么做？我在 MySql 中尝试了以下操作：

SELECT p.person_id, p.gender, p.name, disease_id, drug_id 
   FROM person_t as p 
   INNER JOIN disease_t on disease_t.person_id = p.person_id 
   RIGHT OUTER JOIN drug_t on drug_t.person_id = p.person_id 
   WHERE disease_id= 52 AND drug_id != 34;

这为我提供了一个人没有服用 drug_id 34 的所有记录，而不是没有服用 drug_id 34 的人。我将如何编写我想要的查询？

Answer 1

您可以使用 NOT IN：

SELECT p.person_id, p.gender, p.name, disease_id
FROM person_t as p 
INNER JOIN disease_t d on disease_t.person_id = p.person_id 
WHERE disease_id = 52
AND p.person_id NOT IN (SELECT person_id IN drug_t WHERE drug_id = 34)

Answer 2

对于那些不想使用子查询的人：

   SELECT p.person_id, p.gender, p.name, disease_id
     FROM PERSON_T p 
     JOIN DISEASE_T d ON d.person_id = p.person_id 
LEFT JOIN DRUG_T dt ON dt.person_id = p.person_id
                   AND dt.drug_id = 34
    WHERE disease_id = 52
      AND dt.person_id IS NULL

Answer 3

取决于优化器，NOT EXISTS 可能比 NOT IN 更有效。两种方法都试一下，看看哪一种效果最好。

SELECT p.person_id, p.gender, p.name, disease_id, drug_id 
   FROM person_t as p 
   INNER JOIN disease_t on disease_t.person_id = p.person_id 
   WHERE disease_id= 52 AND NOT EXISTS (
       SELECT * from drug_T WHERE person_id = person_t.person_id AND drug_id = 34)