无法进行合适的查询

Question

我正在尝试创建一个数据库。它具有以下表格子集

职位（job_id，描述）
技能（skill_id，技能）
jobs_and_skills (job_id, skill_id)
applicant_skill (applicant_id , skills_id)

我的查询必须是：查找具有给定工作 ID 所需技能的求职者

我使用一系列子查询做了类似的事情

select DISTINCT applicant_id
from applicant_skill
where  skill_id in
(select DISTINCT skill_id
from jobs_and_skill
where  job_id in (select job_id
from jobs
where description ="research"));

但是在这里，如果我有一份需要技能的工作，比如 A、B、C，并且我有候选人只有技能 A，那么 return 那个候选人也会。有没有办法找到具备所有技能 A、B、C 的候选人，而不是 return 只具备所需技能子集的候选人

Answer 1

使用 JOIN 而不是 IN。然后你可以加入一个子查询，计算一份工作所需的技能数量，并将其与申请人共有的技能数量进行比较。

SELECT as.applicant_id
FROM applicant_skill AS as
JOIN jobs_and_skills AS js ON js.skill_id = as.skill_id
JOIN jobs AS j ON j.job_id = js.job_id
JOIN (
    SELECT job_id, COUNT(*) AS skill_count
    GROUP BY job_id
) AS jsc ON j.job_id = jsc.job_id
WHERE j.description = "research"
GROUP BY as.applicant_id
HAVING COUNT(*) = jsc.skill_count

Answer 2

您可以在加入 applicants 之前为 jobs_and_skill 准备 subquery。

select * from applicant_skill t1
inner join 
    (select t2.skill_id from 
        jobs_and_skill t2
        inner join skills t3 on t3.skill_id = t2.skill_id
        inner join jobs t4 on t4.job_id = t2.job_id
        where t4.description  = 'research') as t 
    on t1.skill_id = t.skill_id

Answer 3

您可以加入jobs和applicants，条件是确保申请人具备所有工作技能。

select 
    j.*,
    a.*
from 
    jobs j
    inner join applicants a on not exists (
        select 1
        from job_and_skills js
        left join applicant_skills as
            on  as.applicant_id = a.applicant_id
            and as.skill_id = js.skill_id
        where 
            js.job_id = j.job_id
            and as.applicant_id is null
    )
where j.description = 'research'

连接条件使用 not exists 条件使用左连接反模式，其短语如下：申请人不拥有此职位所需的技能。

无法进行合适的查询

Unable to make a suitable query

mysql

database

join

count

sql-in