oracle查詢根據條件刪除重複行

Question

我有類似下表的數據。

+------+----------+----------+ 
| 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 | 
+------+----------+----------+ 

我需要通過除去重複項以獲得行，結果是下面

+------+----------+----------+ 
| 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 | 
+------+----------+----------+  

用於particaular類型，

1. 如果相應的日期2存在對DATE1，然後選擇該行並刪除其他日期1
2. 如果相應的日期2不存在，則保留它

同樣，

如果相應DATE1存在對DATE2，然後選擇該行，並刪除其它DATE2

如果相應的日期1不存在，那麼保持它

是否有可能爲此編寫oracle查詢？ 任何人都可以幫助我做到這一點？

Answer 1

嘗試此查詢：

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