Question

我正在尝试查询视图，但得到＆＃34; ORA-01873：间隔的前导精度太小＆＃34;错误信息。以下是查询。

Select * from table order by Col1.

以下是视图结构：

Col1    NOT NULL NUMBER
Col2    NOT NULL NVARCHAR2(80)
Col3    NOT NULL NUMBER
Col4    NOT NULL VARCHAR2(10)
Col5        NVARCHAR2(80)
Col6        NVARCHAR2(255)
Col7        NUMBER
Col8    NOT NULL NVARCHAR2(255)
Col9    NOT NULL NVARCHAR2(1)
Col10   NOT NULL NUMBER
Col11       VARCHAR2(19)
Col12       VARCHAR2(19)
Col13       VARCHAR2(19)
Col14       VARCHAR2(19)
Col15       VARCHAR2(19)
Col16       VARCHAR2(19)
Col17       NUMBER
Col18       NVARCHAR2(255)
Col19       NVARCHAR2(80)
Col20   NOT NULL NUMBER

以下是视图定义：

SELECT tab2.cola AS Col1,
    tab1.col AS COl2,    
    tab2.colb AS Col3,
    tab7.col AS Col4,
    DECODE(tab3.col, NULL, tab4.col) AS COl5,
    tab8.col AS Col6,
    tab6.col AS COl7,
    tab2.Colc AS Col8,
    tab2.cold AS Col9,
    tab2.cole AS Col10,
    TO_CHAR(TO_DATE('1970/01/01 00:00:00', 'YYYY/MM/DD HH24:MI:SS') + NUMTODSINTERVAL( tab2.colf / 1000,'SECOND'), 'YYYY/MM/DD HH24:MI:SS') AS COl11,
    TO_CHAR(TO_DATE('1970/01/01 00:00:00', 'YYYY/MM/DD HH24:MI:SS') + NUMTODSINTERVAL( tab2.colg / 1000,'SECOND'), 'YYYY/MM/DD HH24:MI:SS') AS Col12,
    TO_CHAR(TO_DATE('1970/01/01 00:00:00', 'YYYY/MM/DD HH24:MI:SS') + NUMTODSINTERVAL( tab2.colh / 1000,'SECOND'), 'YYYY/MM/DD HH24:MI:SS') AS Col13,
    TO_CHAR(TO_DATE('1970/01/01 00:00:00', 'YYYY/MM/DD HH24:MI:SS') + NUMTODSINTERVAL( tab2.coli / 1000,'SECOND'), 'YYYY/MM/DD HH24:MI:SS') AS Col14,
    TO_CHAR(TO_DATE('1970/01/01 00:00:00', 'YYYY/MM/DD HH24:MI:SS') + NUMTODSINTERVAL( tab2.colj / 1000,'SECOND'), 'YYYY/MM/DD HH24:MI:SS') AS COl15,
    TO_CHAR(TO_DATE('1970/01/01 00:00:00', 'YYYY/MM/DD HH24:MI:SS') + NUMTODSINTERVAL( tab2.colk / 1000,'SECOND'), 'YYYY/MM/DD HH24:MI:SS') AS COl16,
    tab2.coll AS Col17,
    tab9.col AS Col18,
    tab10.col AS Col19,
    tab2.colm AS Col20
FROM tab1 ,
    tab2 ,
    tab3,
    tab4,
    tab5,
    tab6,
    tab7,
    tab8,
    tab9,
    tab10
WHERE ....

其他查询运行正常，例如select * from table where Col1 = 123或select * from table where Col2 = 'xyz' order by Col1，但上述查询不适用于任何col。请建议。

Answer 1

您的numtodsinterval()函数中有一个数字'epoch'数字似乎太大（或太小）无法处理。您可以传递的最大值为秒数^ 2 31-1：

SQL> select numtodsinterval(power(2,31) - 1, 'SECOND') as interval from dual; 

INTERVAL     
--------------
24855 3:14:7.0

SQL> select numtodsinterval(power(2,31), 'SECOND') as interval from dual; 

SQL Error: ORA-01873: the leading precision of the interval is too small
01873. 00000 -  "the leading precision of the interval is too small"
*Cause:    The leading precision of the interval is too small to store the
           specified interval.
*Action:   Increase the leading precision of the interval or specify an
           interval with a smaller leading precision.

作为一个时代，允许的最高秒数代表2038-01-19 03:14:07。 This is the year 2038 problem，基本上。

您也可以使用负数来到达那里：

SQL> select numtodsinterval(-2208988800, 'SECOND') as interval from dual;

SQL Error: ORA-01873: the leading precision of the interval is too small

使用-power(2, 31)换行为正值，但低于该值的任何值：

SQL> select numtodsinterval(power(2,31) - 1, 'SECOND') as interval from dual;

INTERVAL     
--------------
24855 3:14:7.0

SQL> select numtodsinterval(-power(2,31), 'SECOND') as interval from dual;

INTERVAL     
--------------
24855 3:14:8.0

SQL> select numtodsinterval(-power(2,31) - 1, 'SECOND') as interval from dual;

SQL Error: ORA-01873: the leading precision of the interval is too small

您要除以1000，因此您的列F到K之一的值超过2147483647000.这应该相当容易找到，您可能要考虑向这些列添加检查约束，以便它们不能设置得太高 - 检查列值是否小于或等于1000 * (power(2, 31) - 1)。并且要么大于零，要么大于 - 1000 * (power(2, 31)。

当你有一个像where Col1 = 123这样的过滤器时没有错误的原因是你的过滤器（谓词）被推到视图查询中，并且不会评估值太高的行。也许您只有一个这样的值，其col1值不 123且其col2值不'xyz'。如果您确定问题行并使用其实际col1值进行过滤，则仍会出错。如果没有过滤器，则会对所有行进行评估。

您拥有的具体负数似乎是一个神奇的数字：

SQL> select date '1970-01-01' - 2208988800/86400 from dual;

DATE'1970-01-01'-2208988800/86400
---------------------------------
1900-01-01 00:00:00

如果要排除，那么您可以修改视图定义以添加过滤器，例如：

...
AND tab2.colh > 0

或更改列表达式来处理它，通过对它进行rignoring并将其保留为null，或者可能更有用地返回该魔术日期：

    TO_CHAR(CASE WHEN tab2.colh = -2208988800000 THEN DATE'1900-01-01'
      ELSE DATE'1970-01-01' + NUMTODSINTERVAL( tab2.colh / 1000,'SECOND')
      END, 'YYYY/MM/DD HH24:MI:SS') AS Col13,

您也可以使用间隔更改为使用日期算术：

    TO_CHAR(DATE'1970-01-01' + ( tab2.colh / 86400000 ), 'YYYY/MM/DD HH24:MI:SS') AS Col13,

你必须修改视图定义而不是你的查询，除非colh包含在选择列表中（它似乎不是），即使它只是你可以排除它 - 并且仍然可能并不总是避免错误，具体取决于优化器处理查询的方式。

ORA-01873：领先的精度

1 个答案: