Question

我有两张桌子：

t_job
 - id
 - name

t_person:
 - id
 - name
 - job_id (fk t_job->id)

t_other_jobs
 - person_id (fk t_person->id)
 - job_id (fk t_job->id)

所以基本上我有人有主要工作和0个或更多的二级工作。我想要做的是创建查询，给我这样的东西：

PERSON NAME | ALL JOBS COUNT

和

JOB NAME | HOW MANY PEOPLE HAVE THEM

因此，如果一个人有主要工作（它是一个外键，所以他必须拥有它）和另外两个工作。结果将是：

Jorge | 3

Answer 1

既然你说他们总是有工作，你可以算上他们的其他工作并加1：

SELECT
  p.id,
  p.name,
  count(o.job_id)+1 as Jobs
FROM
  t_person p
LEFT JOIN 
  t_other_jobs o
ON
  p.id=o.person_id
GROUP BY
  p.id, p.name;

Answer 2

这个qwery会给你第一个结果。

select 
    person.id,
    count(distinct (person.job_id)) + count(other_jobs.job_id) as All_job
from
    person
        Inner JOin
    other_jobs
where
    person.id = other_job.person_id
group by person.id

这将为您提供第二个。

select 
    job.name, count(distinct (person.id)) as no_of_people
from
    person
        INNER JOIN
    job ON person.job_id = job.id
group by job_id 
UNION select 
    job.name, count(distinct (person_id)) as no_of_people
from
    other_job
        INNER join
    job ON job.id = other_job.job_id
group by job_id;

Answer 3

尝试

第一次查询（PERSON NAME | ALL JOBS COUNT）

SELECT name PersonName, maincount + Count(o.Job_Id) AS AllJobsCount
FROM (
  SELECT p.Id, p.name, COUNT(m.Id) AS MainCount
  FROM  t_person p
  LEFT JOIN t_job m ON  p.job_id = m.Id
  GROUP BY P.Id
) a
Left Join t_other_jobs o ON A.Id = o.person_id
GROUP BY a.Id

第二次查询（JOB NAME | HOW MANY PEOPLE HAVE THEM）

SELECT j.Name JobName, 
       (SELECT COUNT(1) FROM t_person WHERE job_id=j.Id) +
       (SELECT COUNT(1) FROM t_other_jobs WHERE job_id=j.Id) AS Count
FROM t_job j

MySql：来自两个选择的count（*）

3 个答案: