甲骨文双重选择问题

CLIENT

cl_id cl_name 
1  John 
2  Maria

付款

pa_id pa_date  pa_status cl_id 
1  2017-01-01  1   1 
2  2017-01-01  1   2 
3  2017-02-01  1   1 
4  2017-02-01  1   2 
5  2017-03-01  0   1 
6  2017-03-01  1   2

我需要一个选择众所周知声明，让我的客户ID，姓名和他的最后付款的状态。所以，我选择的最终结果应该是：

cl_id cl_name pa_status 
1  John  0 
2  Maria 1

这是客户选择的作品：

select cl_id, cl_name from CLIENT;

就是付款选择工作的最后状态：

select * from ( select pa_status from PAYMENT ORDER BY PA_DATE DESC) where rownum = 1;

所以现在，我需要让他们一起工作。我试过2种方法没有奏效：

select cl_id, cl_name, (select * from ( select pa_status from PAYMENT ORDER BY PA_DATE DESC) where rownum = 1 and PAYMENT.cl_id = CLIENT.CL_ID) as last_status from CLIENT;

错误：无效的标识符

这：

select cl_id, cl_name, (select * from ( select pa_status from PAYMENT ORDER BY PA_DATE DESC) where rownum = 1) as last_status from CLIENT;

不给我任何错误，但只显示约翰的最后一个状态，即最后一条记录：

cl_id cl_name last_status 
1  John  0 
2  Maria 0

任何人都可以给我一个提示吗？

感谢

来源

2017-04-12 Sambarilove

您需要使用分析功能。这种功能可以让你将数据分割成一些组，并按照你的意愿对每个组的数据进行排名。

你的情况：

Select * from (
Select id, name, status, row_number() over (partition by  p.cl_id order by p.pa_date desc) as rw 
From client c join payments p on p.cl_id = c.cl_id) 
Inn where inn.rw = 1;

来源

2017-04-12 18:09:16 user2671057

这工作很好的家伙。谢谢！ – Sambarilove

太好了。还有更好的语法 - 寻找FIRST分析函数。祝你好运！ – user2671057

这得到了客户最大日期，然后获得最高支付编号与日期。

with max_date as (
     select max(date) as max_date, cl_id from payments group by cl_id 
) 
select c.cl_id, c.cl_name, p.pa_sttus from client c 
join payments p 
on c.cl_id = p.cl_id 
where p.pa_id = (select max(p2.pa_id) from payments p2 
        join max_date md 
        on p2.cl_id = md.cl_id 
        where p.cl_id = p2.cl_id 
        and p2.pa_date = md.max_date 
       )

来源

2017-04-12 17:26:22 gorrilla10101

这给了我最后一行无效的标识符：“md.pa_date” – Sambarilove

我使用了错误的列名。应该现在工作。 – gorrilla10101

先取日期的最大距离为每个客户端ID。

Select cl_id, max(pa_date) as pa_date from PAYMENTS group by cl_id

现在你把乌尔客户端表，并与上述子查询加入

select c.cl_id, c.cl_name, 
(select pa_status from PAYMENT t where t.pa_date=p.pa_date and t.cl_id=p.cl_id) 
from CLIENT c join (Select cl_id, max(pa_date) as pa_date from PAYMENTS group by cl_id) p on p.cl_id=c.cl_id

来源

2017-04-12 18:22:40 Rams

您可以使用Oracle的KEEP LAST这里：

select cl_id, c.cl_name, last_payment.status 
from client 
join 
(
    select 
    cl_id, 
    max(pa_status) keep (dense_rank last order by pa_date) as status 
    from payments 
    group by cl_id 
) last_payment using (cl_id);

（如果要包括但不付款的客户，变化加入到LEFT OUTER JOIN。）

来源

2017-04-13 08:21:00

甲骨文双重选择问题

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