三-table 加入 NO EXISTS

Question

我有三个 table，想在 SQL 中回答以下问题：

“谁只有证书没有分数？”

例如，在下面的设置中，查询只会 return “John”。 Joana 拥有 SCORE table (id 57) 中的“AWS 认证”。 Marry 拥有不在 SCORE table 中的“ITIL V3 认证”，但她也有“Professional Python Dev”，在你的 SCORE table.

中

PERSON
| PERSON_ID | PERSON_NAME |
| --------  | -------------- |
| 12    | John            |
| 23   | Mary            |
| 24   | Joana            |


**CERTIFICATION**
|CERTIFICATION_ID| CERTIFICATION_NAME | PERSON_ID|
| -------- |----------- | -------------- |
| 53 | Java Certification    |    12       |
| 54 | ITIL V3 Certification    |    23       |
| 55 | Professional Python Dev            | 23|
| 56 |GCP Certification            |23   | 
| 57 |AWS Certification            |24   | 

SCORES

|SCORE_ID| CERFITICATION_ID | SCORE_DETAILS |
| -------- |----------- | -------------- |
| 70 |55   | 80%            |
| 71 |56   | 90%            |
| 72 |57   | 95%           |

我试图在 SQL 中实现它，而不必遍历记录并且不使用存储过程。

SQL 用于创建这些 table 并添加数据以备不时之需：

create table person(
person_id integer,
person_name varchar(16)
);

create table certification(
CERTIFICATION_ID int,
  CERTIFICATION_NAME varchar(16),
  person_id int
);

create table scores(
  SCORE_ID int,
  CERTIFICATION_ID int,
 SCORE_DETAILS varchar(16));


insert into person(person_id, person_name) values(12, 'John');
insert into person(person_id, person_name) values(23, 'Mary');
insert into person(person_id, person_name) values(24, 'Joana');


insert into certification(CERTIFICATION_ID,CERTIFICATION_NAME,person_id) values(53,'A', 12);
insert into certification(CERTIFICATION_ID,CERTIFICATION_NAME,person_id) values(54,'B',23);
insert into certification(CERTIFICATION_ID,CERTIFICATION_NAME,person_id) values(55,'C', 23);
insert into certification(CERTIFICATION_ID,CERTIFICATION_NAME,person_id) values(56,'D', 23);
insert into certification(CERTIFICATION_ID,CERTIFICATION_NAME,person_id) values(57,'E', 24);

insert into scores (SCORE_ID,CERTIFICATION_ID, SCORE_DETAILS) values (70,55,'e');
insert into scores (SCORE_ID,CERTIFICATION_ID, SCORE_DETAILS) values (71,56,'f');
insert into scores (SCORE_ID,CERTIFICATION_ID, SCORE_DETAILS) values (72,57,'g');

Answer 1

我们可以像下面这样使用not

select p.* from person p
 where p.person_id not in (
select c.person_id from 
certification c join scores s on c.CERTIFICATION_ID=s.CERTIFICATION_ID
)

demo link

Answer 2

您可以使用以下查询

select *
  from person p
 where not exists (select 1
                 from certification c
                 join scores s on s.certification_id = c.certification_id
                where p.person_id = c.person_id
              )

三-table 加入 NO EXISTS

Three-table join with NO EXISTS

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