Question

我想获得以下结果。请帮助Oracle SQL查询。

结果：


ID | APP_NUMBER | DATE
1  | ABC4100 | 10/02/2016
11 | ABC1001 | 10/04/2016
34 | ABC2001 | 9/15/2016

上述结果来自以下两个表：

表1：


ROW | ID | APP_NUMBER | DATE
1   | 1  | ABC4000    | 01/01/2016
2   | 1  | ABC4100    | 01/02/2016
3   | 1  | ABC4100    | 01/03/2016
4   | 11 | ABC1001    | 04/01/2016
5   | 34 | ABC2001    | 05/01/2016

表2：


APP_NUMBER | DATE
ABC4000    | 01/01/2016
ABC4100    | 08/02/2016
ABC4100    | 09/15/2016
ABC4100    | 10/02/2016
ABC1001    | 10/01/2016
ABC1001    | 10/02/2016
ABC1001    | 10/03/2016
ABC1001    | 10/04/2016
ABC2001    | 09/14/2016
ABC2001    | 09/15/2016

Answer 1

试试这个：

select id,
       app_number,
       "date" 
from (
      select distinct table_1.id,
            table_1.app_number,
            table_2."date" ,
            rank() over (partition by id order by table_2."date" desc) r
from  table_1,
      table_2
where table_1.APP_NUMBER=table_2.APP_NUMBER
)
where r=1
order by id;

Answer 2

这将为您提供所需的结果。

with table2(APP_NUMBER, DT) as (  select 'ABC4000',to_date ('01/01/2016','mm/dd/yyyy') from dual
                                  UNION ALL
                                  select 'ABC4100',to_date ('08/02/2016','mm/dd/yyyy') from dual
                                  UNION ALL
                                  select 'ABC4100',to_date ('09/15/2016','mm/dd/yyyy') from dual
                                  UNION ALL
                                  select 'ABC4100',to_date ( '10/02/2016','mm/dd/yyyy') from dual
                                  UNION ALL
                                  select 'ABC1001',to_date ( '10/01/2016','mm/dd/yyyy') from dual
                                  UNION ALL
                                  select 'ABC1001',to_date ('10/02/2016','mm/dd/yyyy') from dual
                                  UNION ALL
                                  select 'ABC1001',to_date ( '10/03/2016','mm/dd/yyyy') from dual
                                  UNION ALL
                                  select 'ABC1001',to_date ('10/04/2016','mm/dd/yyyy') from dual
                                  UNION ALL
                                  select 'ABC2001',to_date ( '09/14/2016','mm/dd/yyyy') from dual
                                  UNION ALL
                                  select 'ABC2001',to_date ( '09/15/2016','mm/dd/yyyy') from dual ),
       table1(ROWN,ID,APP_NUMBER,DTE) as ( select '1', '1','ABC4000',to_date( '01/01/2016','mm/dd/yyyy') from dual 
                                           UNION ALL
                                           select '2', '1','ABC4100',to_date( '01/02/2016','mm/dd/yyyy') from dual 
                                           UNION ALL
                                           select '3', '1','ABC4100',to_date( '01/03/2016','mm/dd/yyyy') from dual 
                                           UNION ALL
                                           select '4', '11','ABC1001',to_date( '04/01/2016','mm/dd/yyyy') from dual 
                                           UNION ALL
                                           select '5', '34','ABC2001',to_date( '05/01/2016','mm/dd/yyyy') from dual 
                                          )

select max(t1.id),t2.app_number,max(t2.dt)
from    table2 t2
inner join  table1 t1
ON t1.APP_NUMBER = t2.APP_NUMBER
where t1.APP_NUMBER <> 'ABC4000'
group by  t2.APP_NUMBER ,t1.id      
order by t1.id  ;

Answer 3

SELECT t1.ID, t1.APP_NUMBER, MAX(t2.Date)
FROM Table1 t1
INNER JOIN Table2 t2
ON t1.APP_NUMBER = t2.APP_NUMBER
WHERE t1.APP_NUMBER IN ('ABC4100', 'ABC1001', 'ABC2001')
ORDER BY t1.ID

这应该有效，但是如果它没有，你可能想尝试t2.APP_NUMBER。

Answer 4

试试这个 -

SELECT ID,t1.APP_NUMBER,MAX(t2.Date)
FROM Table1 t1
INNER JOIN Table2 t2
ON t1.APP_NUMBER=t2.APP_NUMBER
GROUP BY ID,t1.APP_NUMBER

Answer 5

您可以使用LEFT JOIN连接两个表中的结果，如下例所示：

Select * from Table1 t1 LEFT JOIN Table2 t2 ON (t1.APP_NUMBER = t2.APP_NUMBER);

如何使用或不使用连接从两个表中获取结果？

5 个答案: