如果其他表中不存在id，则获取不同的行并使用合并

d_unknown 发表于 Dev

d_unknown

我有两个表：

Job_Order

-id
-creation_date
-assigned_to
-job_ordertype
-client

Job_Order_Stage

-id
-stage
-date 
-job_order(foreign key to job_order)

job_order如果ID在Job_Order_Stage中不存在，我想获取中的所有行并将阶段设置为0。而且我也只想获取在中找到max(stage)相同的行。我该怎么做？job_orderjob_order_stage

我在这里查询：

SELECT 
    a.id,a.creation_date,
    e.user_name,c.operation,
    c.system_,d.name,
    Coalesce((s.stage), 0) as stage_name
FROM job_order a
INNER JOIN account b ON a.assigned_to=b.id
INNER JOIN job_order_type c ON a.job_order_type=c.id
INNER JOIN user e ON b.user=e.user_id
INNER JOIN client d ON a.client=d.id
LEFT JOIN job_order_stage s ON s.job_order = a.id

我的这个sql语句的问题是，它显示所有作业订单及其重复的副本，并具有不同的阶段。如何解决这个问题？

帕拉

这将获取所有不在job_order_stage中的job_orders

select job_order.*
  from job_order
    left join job_order_stage
      on job_order.id = job_order_stage.job_order
  where job_order_stage.job_order is null;

这可以让您max(stage)找到工作在job_order_stage中的位置

select job_order.id, max(job_order_stage.stage)
  from job_order 
    inner join job_order_stage
      on job_order.id = job_order_stage.job_order
  group by job_order.id;

还是您想以某种方式将它们合并？这将是这样的：

select job_order.*, max(coalesce(job_order_stage.stage, 0)) stage
  from job_order
    left join job_order_stage
      on job_order.id = job_order_stage.job_order
  group by job_order.id

这可以利用/滥用mysqls的特殊处理group by，但是在这种情况下应该没问题。

从评论更新

select *
  from (
    select job_order.*, max(coalesce(job_order_stage.stage, 0)) stage
      from job_order
        left join job_order_stage
          on job_order.id = job_order_stage.job_order
      group by job_order.id
 ) q 
 where stage = 2;

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