Sql oracle识别列的每个更改

我有下表：

CC   | BEGIN               | END                 | ORDER

1215 | 20.03.2015 14:59:21 | 23.03.2015 13:47:19 | 1
1310 | 23.03.2015 13:47:19 | 27.03.2015 19:19:23 | 2
1105 | 27.03.2015 19:19:23 | 03.04.2015 18:10:22 | 3
1205 | 03.04.2015 18:10:22 | 04.04.2015 17:48:51 | 4
4650 | 04.04.2015 17:48:51 | 05.04.2015 13:28:20 | 5
1215 | 05.04.2015 13:28:20 | 10.04.2015 11:00:01 | 6
1310 | 12.04.2015 15:10:52 | 20.04.2015 09:44:39 | 7
1220 | 20.04.2015 09:44:39 | 22.04.2015 18:10:13 | 8
1220 | 23.04.2015 10:46:25 | 23.04.2015 10:49:08 | 9
1215 | 23.04.2015 10:49:08 | 23.04.2015 11:00:14 | 10
1215 | 23.04.2015 11:00:14 | 23.04.2015 11:47:45 | 11
1215 | 23.04.2015 11:47:45 | 29.04.2015 04:33:27 | 12
1215 | 29.04.2015 04:33:27 | 01.05.2015 16:46:53 | 13
2880 | 01.05.2015 16:46:53 | 06.05.2015 15:55:13 | 14
1210 | 06.05.2015 15:55:13 | 08.05.2015 16:35:40 | 15
1210 | 08.05.2015 16:35:40 | 15.05.2015 18:10:00 | 16
1205 | 17.05.2015 14:19:01 | 21.05.2015 15:29:08 | 17
1205 | 21.05.2015 15:29:08 | 24.05.2015 16:59:14 | 18
1205 | 24.05.2015 16:59:14 | 27.05.2015 14:33:59 | 19
1205 | 27.05.2015 14:33:59 | 01.06.2015 12:42:45 | 20
1205 | 27.05.2015 14:33:59 | 01.06.2015 12:42:45 | 20

我正在寻找一种方法，每次列CC更改时都输入一个唯一的数字。 例如：

---

我尝试过使用LAG（）和LEAD（）函数，它只在CC列重复两次时工作正常这是通常使用

秩
/
稠密的_秩
函数所做的

请看，您需要以某种方式使用它：

select ..., dense_rank() over (order by cc) column_i_wish, ...

这应该起作用：

SELECT CC, "BEGIN", "END", "ORDER",
       SUM(flag) OVER (ORDER BY "ORDER")  AS new_column
FROM (
  SELECT CC, "BEGIN", "END", "ORDER",
         CASE WHEN LAG(CC) OVER (ORDER BY "ORDER") = CC THEN 0 ELSE 1 END AS flag
  FROM mytable) AS t

这不起作用，因为订单是由
“order”
字段指定的，而不是
“CC”
字段指定的。
SELECT CC, "BEGIN", "END", "ORDER",
       SUM(flag) OVER (ORDER BY "ORDER")  AS new_column
FROM (
  SELECT CC, "BEGIN", "END", "ORDER",
         CASE WHEN LAG(CC) OVER (ORDER BY "ORDER") = CC THEN 0 ELSE 1 END AS flag
  FROM mytable) AS t