如何根据左连接条件获取 mysql 值

Question

工作table

1.id
2.status = 'active'
3.name

修复table

1.repair id
2.job_id
3.process = 'yes|no'
4.status  = '1|2'

工作table

id   name  status
1    test  active
2    check active

修复table

repair_id     job_id  process  status
1                1      no        2
2                1      no        1
3                1      yes       2
4                2      no        1
5                2      no        2

这里我需要显示数据，其中（进程！= 'yes' 和 repair_status ！= 2）按 job_id

分组

我需要查询后的结果

---------------------------------------------
job_id    name( job.name ) status( job.status )
------------------------------------------------
2            check           active

Answer 1

如果您想要 process = 'yes' 和 status = 2 没有修复的工作，您可以使用 not exists:

select j.*
from jobs j
where not exists (
    select 1 
    from repair r 
    where r.job_id = j.id and r.process = 'yes' and r.status = 2
)

Answer 2

为了获得您指定的结果，您的意思是 process 不是 yes 对于任何行对于 job_id .然后至少一行有 status <> 2。那将是：

select j.job_id, j.name, j.status
from repair r join
     job j
     on r.job_id = r.id
group by j.job_id, j.name, j.status
having max(process) = 'no' and
       min(repair_status) = 1;

如何根据左连接条件获取 mysql 值

how to get mysql values based on left join condition

mysql

sql

subquery

left-join

having-clause