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sql - 使用 SQL 在数字范围之间生成后续数字

转载 作者:行者123 更新时间:2023-12-05 01:48:21 24 4
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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,其中我列出了定义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 & model 子句,但到目前为止没有成功。

谢谢你的帮助

最佳答案

使用示范条款:

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>

关于sql - 使用 SQL 在数字范围之间生成后续数字,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/14120920/

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