oracle查询根据条件删除重复行

Question

我有类似下表的数据。

+------+----------+----------+ 
| type | date1 | date2 | 
+------+----------+----------+ 
| IT1 | 05/01/15 | 08/01/15 | 
| IT1 | 05/01/15 |   | 
| IT1 | 04/01/15 |   | 
| IT1 |   | 03/02/15 | 
| IT1 | 06/01/15 | 03/02/15 | 
| IT1 |   | 04/02/15 | 
| IT2 | 05/01/15 |   | 
| IT2 | 05/01/15 | 04/01/15 | 
| IT2 | 03/01/15 |   | 
| IT2 |   | 09/01/15 | 
+------+----------+----------+ 

我需要通过除去重复项以获得行，结果是下面

+------+----------+----------+ 
| type | date1 | date2 | 
+------+----------+----------+ 
| IT1 | 05/01/15 | 08/01/15 | 
| IT1 | 04/01/15 |   | 
| IT1 | 06/01/15 | 03/02/15 | 
| IT1 |   | 04/02/15 | 
| IT2 | 05/01/15 | 04/01/15 | 
| IT2 | 03/01/15 |   | 
| IT2 |   | 09/01/15 | 
+------+----------+----------+  

用于particaular类型，

1. 如果相应的日期2存在对DATE1，然后选择该行并删除其他日期1
2. 如果相应的日期2不存在，则保留它

同样，

如果相应DATE1存在对DATE2，然后选择该行，并删除其它DATE2

如果相应的日期1不存在，那么保持它

是否有可能为此编写oracle查询？ 任何人都可以帮助我做到这一点？

Answer 1

尝试此查询：

select * 
from table1 t1 
WHERE 
    "date2" IS NOT NULL 
    AND 
    "date1" IS NOT NULL 
    OR 
    "date2" IS NULL 
    AND NOT EXISTS (
    SELECT 1 FROM table1 t2 
    WHERE t1."type" = t2."type" 
     AND t1."date1" = t2."date1" 
     AND t2."date2" IS NOT NULL 
    ) 
    OR 
    "date1" IS NULL 
    AND NOT EXISTS(
    SELECT 1 FROM table1 t2 
    WHERE t1."type" = t2."type" 
     AND t1."date2" = t2."date2" 
     AND t2."date1" IS NOT NULL 
    ) 

演示：http://sqlfiddle.com/#!4/12b8b/8