2

我有一个包含以下内容的表格：

ID    low_value     high_value
1     3270200000    3270210000
2     3270210000    3270220000
3     3270220000    3270230000
4     3270230000    3270231000
5     3270231000    3270232000
6     3270232000    3270240000
...

在使用 Oracle 11g 的单个查询中，我想检索以下结果，其中我列出了存在于定义的范围内的每个唯一数字

start value = with low_value /
end value = high_value - 1

 

low_value   high_value    unique_value
3270200000  3270210000    3270200000    
3270200000  3270210000    3270200001
3270200000  3270210000    3270200002
3270200000  3270210000    3270200003
...
3270200000  3270210000    3270209999
3270210000  3270220000    3270210001
3270210000  3270220000    3270210002
3270210000  3270220000    3270210002
...
3270210000  3270220000    3270219999
...

我一直在玩 connect by & 模型子句，但到目前为止还没有成功。

TX 为您提供帮助

4

1 回答 1

5

使用模型子句：

SQL> create table foo1
  2  (id number,
  3  low_value number,
  4  high_value number);

Table created.

SQL>
SQL> insert into foo1 values (1,  3270200000, 3270200010);

1 row created.

SQL> insert into foo1 values (2,  3270210000, 3270210005);

1 row created.

SQL> insert into foo1 values (3,  10, 10);

1 row created.

SQL> commit;

Commit complete.

SQL>
SQL>
SQL> with foo as
  2   (select f.id, f.low_value, f.high_value, f.high_value - f.low_value
  3     range from foo1 f)
  4  select key, low_value, high_value, unique_value
  5    from foo
  6   model partition by(id as key)
  7         dimension by(0 as f)
  8         measures(low_value as unique_value, low_value, high_value, range)
  9         rules (unique_value [for f from 0 to range[0]  increment 1]  = low_value[0] + cv(f),
 10               low_value[for f from 0 to range[0]  increment 1] = low_value[0],
 11               high_value[for f from 0 to range[0]  increment 1] = high_value[0]);

       KEY  LOW_VALUE HIGH_VALUE UNIQUE_VALUE
---------- ---------- ---------- ------------
         1 3270200000 3270200010   3270200000
         1 3270200000 3270200010   3270200001
         1 3270200000 3270200010   3270200002
         1 3270200000 3270200010   3270200003
         1 3270200000 3270200010   3270200004
         1 3270200000 3270200010   3270200005
         1 3270200000 3270200010   3270200006
         1 3270200000 3270200010   3270200007
         1 3270200000 3270200010   3270200008
         1 3270200000 3270200010   3270200009
         1 3270200000 3270200010   3270200010
         2 3270210000 3270210005   3270210000
         2 3270210000 3270210005   3270210001
         2 3270210000 3270210005   3270210002
         2 3270210000 3270210005   3270210003
         2 3270210000 3270210005   3270210004
         2 3270210000 3270210005   3270210005
         3         10         10           10

18 rows selected.

SQL>

和 11g 递归因式分解

SQL> with foo (id, low_value, high_value, unique_value)
  2        as (select f.id, f.low_value, f.high_value, low_value unique_value
  3              from foo1 f
  4             union all
  5            select id, low_Value, high_value, unique_value + 1
  6              from foo
  7             where unique_value < high_value)
  8  select id, low_value, high_value, unique_value
  9    from foo
 10   order by id, unique_value
 11  /

        ID  LOW_VALUE HIGH_VALUE UNIQUE_VALUE
---------- ---------- ---------- ------------
         1 3270200000 3270200010   3270200000
         1 3270200000 3270200010   3270200001
         1 3270200000 3270200010   3270200002
         1 3270200000 3270200010   3270200003
         1 3270200000 3270200010   3270200004
         1 3270200000 3270200010   3270200005
         1 3270200000 3270200010   3270200006
         1 3270200000 3270200010   3270200007
         1 3270200000 3270200010   3270200008
         1 3270200000 3270200010   3270200009
         1 3270200000 3270200010   3270200010

        ID  LOW_VALUE HIGH_VALUE UNIQUE_VALUE
---------- ---------- ---------- ------------
         2 3270210000 3270210005   3270210000
         2 3270210000 3270210005   3270210001
         2 3270210000 3270210005   3270210002
         2 3270210000 3270210005   3270210003
         2 3270210000 3270210005   3270210004
         2 3270210000 3270210005   3270210005
         3         10         10           10

18 rows selected.

SQL>
于 2013-01-02T11:58:25.360 回答