Left join with distinct 并计算列值

Question

我有两个类似的表，我想加入它们并得到如下所示的结果

Table 1

JobNO       |       JobMovRefNo
---------------------------------
123         |   1
456         |   1
789         |   1

Table 2

JobNO       |   Arrived
---------------------------------
123         |   y
123         |   y
123         |   n
456         |   y
456         |   y

结果Table

JobNO       |   Arrived
---------------------------------
123         |   n
456         |   y
789         |   n

我尝试使用 group by 进行左连接，但我得到了重复的记录。非常感谢任何帮助。

提前致谢。

Answer 1

使用 left join 并在加入前聚合

select *
from table1 t1 left join
     (select jobno, min(arrived) as arrived
      from table2 t2
      group by jobno
     ) t2
     using (jobno);

你没有解释你的逻辑，但你似乎想要每行的最小值 arrived，这就是它的作用。

编辑：

非常旧的 DB2 版本不支持 USING。这也可以写成：

select t1.*, coalesce(t2.arrived, 'n') as arrived
from table1 t1 left join
     (select jobno, min(arrived) as arrived
      from table2 t2
      group by jobno
     ) t2
     on t2.jobno = t1.jobno;

Answer 2

做一个LEFT JOIN，GROUP BY。使用 coalesce 获取 'n' 表 2 中不存在的作业编号。

select t1.jobno, coalesce(min(t2.arrived), 'n')
from table1 t1
left join table2 t2 on t1.jobno = t2.jobno
group by t1.jobno

Answer 3

这是我的解决方案。我使用了 postgreSQL；希望它适用于您的 dbms。

select 
    t1.jobno,
    min(CASE WHEN t2.arrived IS NULL THEN 'n' ELSE t2.arrived END) as arrived
from table1 t1 left join table2 t2 on t1.jobno = t2.jobno  
group by t1.jobno
order by t1.jobno

Left join with distinct 并计算列值

Left join with distinct and calculate column value

c#

sql

db2

group-by

left-join