如何找出oracle数字的位数 [英] How to find out the number of digits of an oracle number

问题描述

我在 oracle 中有一个看起来像这样的表:

I have a table in oracle that looks like this:

name              |     type    | nullable  
------------------------------------------
person_name       | varchar(20) | yes  
weight_coeficient | number      | yes
...

我怎样才能算出weight_coeficient 的值有多少位?例如:

How can I figure out how many digits a value of weight_coeficient has ? For example:

3.0123456789 有 11 位数字(精度 = 11) 和小数点后 10 位数字 (比例 = 10)

3.0123456789 has 11 digits (precision = 11) and 10 digits after the decimal (scale = 10)

是否有执行此操作的 sql 命令/函数，例如返回 11 的 GetPrecision( select.. ) ?

Is there any sql command/function that does that, something like GetPrecision( select.. ) that returns 11 ?

另请注意，表的定义未指定比例和精度.所以据我所知，最大精度适用于所有数字.所以我对找出定义的精度(= 48)不感兴趣，而是对表中特定值的精度感兴趣.是否可以仅使用 oracle 命令?

Note also that the definition of the table does not specify scale and precision. So as far as I know the maximum precision is applied for all the numbers. So I'm not interested in finding out the precision (= 48) of the definition, but the precision of a specific value in the table. Is that possible just using oracle commands ?

提前谢谢你，
JP

Thank you in advance,
JP

推荐答案

怎么样....

SELECT LENGTH(TRANSLATE(TO_CHAR(3.0123456789),'1234567890.-','1234567890')) 
FROM dual

翻译只是删除非数字字符.-

The translate simply removes the non numeric characters .-

这篇关于如何找出oracle数字的位数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！