递归SQL给予ORA-01790

使用Oracle 11g第2版，以下查询给出了一个ORA-01790：表达式必须具有相同的数据类型相应的表达：递归SQL给予ORA-01790

with intervals(time_interval) AS 
(select trunc(systimestamp) 
    from dual 
    union all 
    select (time_interval + numtodsinterval(10, 'Minute')) 
    from intervals 
    where time_interval < systimestamp) 
select time_interval from intervals;

该错误表明的两个子查询的数据类型UNION ALL正在返回不同的数据类型。

即使我在每个子查询中转换到TIMESTAMP，我也会得到相同的错误。

我错过了什么？

编辑：我不想找一个CONNECT BY替换。

来源

2010-04-06 PenFold

什么time_interval'的'类型以及为什么您的标题说*递归*？ – 2010-04-06 17:15:02

@Peter：WITH视图引用自身。这在11gR2中是新的。请参阅http://download.oracle.com/docs/cd/E11882_01/server.112/e10881/chapter1.htm#FEATURENO08835 – PenFold 2010-04-06 17:17:50

无法看到，抱歉。 – 2010-04-06 17:25:41

在我看来，“递归子查询分解”是对于具有日期或时间戳列的查询，在11g R2中被破解。

with test(X) as 
(
    select to_date('2010-01-01','YYYY-MM-DD') from dual 
    union all (
    select (X + 1) from test where X <= to_date('2010-01-10','YYYY-MM-DD') 
) 
) 
select * from test; 

ORA-01790

使用强制转换为数据类型转换：

with test(X) as 
(
    select cast(to_date('2010-01-01','YYYY-MM-DD') as date) from dual 
    union all (
    select (X + 1) from test where X <= to_date('2010-01-10','YYYY-MM-DD') 
) 
) 
select * from test; 

X 
------------------- 
2010-01-01 00:00:00 

1 row selected

铸造日期为日期是帮助，但如果是其他结果？

它会变得更好......

与另一个开始日期试试：

with test(X) as 
(
    select cast(to_date('2007-01-01','YYYY-MM-DD') as DATE) from dual 
    union all (
    select (X + 1) from test where X <= to_date('2011-01-11','YYYY-MM-DD') 
) 
) 
select * from test 
where rownum < 10; -- important! 

X 
------------------- 
2007-01-01 00:00:00 
2006-12-31 00:00:00 
2006-12-30 00:00:00 
2006-12-29 00:00:00 
2006-12-28 00:00:00 
2006-12-27 00:00:00 
2006-12-26 00:00:00 
2006-12-25 00:00:00 
2006-12-24 00:00:00 

9 rows selected

倒数？为什么？

更新14-JAN-2014：作为一种变通方法，使用CTE开始与结束日期向后构建递归CTE，像这样：

with test(X) as 
(
    select cast(to_date('2011-01-20','YYYY-MM-DD') as DATE) as x from dual 
    union all (
    select cast(X - 1 AS DATE) from test 
    where X > to_date('2011-01-01','YYYY-MM-DD') 
) 
) 
select * from test

结果：

|        X | 
|--------------------------------| 
| January, 20 2011 00:00:00+0000 | 
| January, 19 2011 00:00:00+0000 | 
| January, 18 2011 00:00:00+0000 | 
| January, 17 2011 00:00:00+0000 | 
| January, 16 2011 00:00:00+0000 | 
| January, 15 2011 00:00:00+0000 | 
| January, 14 2011 00:00:00+0000 | 
| January, 13 2011 00:00:00+0000 | 
| January, 12 2011 00:00:00+0000 | 
| January, 11 2011 00:00:00+0000 | 
| January, 10 2011 00:00:00+0000 | 
| January, 09 2011 00:00:00+0000 | 
| January, 08 2011 00:00:00+0000 | 
| January, 07 2011 00:00:00+0000 | 
| January, 06 2011 00:00:00+0000 | 
| January, 05 2011 00:00:00+0000 | 
| January, 04 2011 00:00:00+0000 | 
| January, 03 2011 00:00:00+0000 | 
| January, 02 2011 00:00:00+0000 | 
| January, 01 2011 00:00:00+0000 |

测试使用：

Oracle Database 11g Enterprise Edition Release 11.2.0.1.0 - 64bit Production

来源

2010-05-17 17:36:54 Elmar

我不知道类型不匹配的想法，但这里还是要完成我想你想（在10gR2中工作）的替代方法：

select base_time + numtodsinterval(10*(level-1), 'Minute') 
from (select trunc(systimestamp) base_time from dual) 
connect by base_time + numtodsinterval(10*(level-1), 'Minute') < systimestamp

来源

2010-04-06 19:50:18

感谢戴夫，但我正在寻找一种使用更清晰的递归sql语法的解决方案。 – PenFold 2010-04-06 20:01:12

奇数 - wor如果你绕过varchar S和转换（不投）KS：

WITH intervals(time_interval) AS 
    (SELECT to_char(TRUNC(systimestamp)) 
    FROM dual 
    UNION ALL 
    SELECT to_char(to_timestamp(time_interval) + numtodsinterval(10, 'Minute')) 
    FROM intervals 
    WHERE to_timestamp(time_interval) < systimestamp 
) 
SELECT to_timestamp(time_interval) time_interval 
FROM intervals

来源

2011-09-07 21:52:00 Steve

递归SQL给予ORA-01790

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