为什么我的SQL代码无法正确计算小数点后的小数位数？

Question

SET COUNTDECIMALPLACES =  
    CASE Charindex('.', NEWNUMBER)
        WHEN 0 THEN 0
        ELSE
          LEN(
            CAST(
                 CAST(
                    REVERSE( NEWNUMBER ) AS float
                     ) AS bigint
                ) 
           )
    END   

If the number is 222.9375 it will incorrectly say that COUNTDECIMALPLACES = 3. But, if the number is 17.5548 it will correctly say that COUNTDECIMALPLACES = 4. Anybody know why?

EDIT 1: TO answer the questions in the comments I am using SQL Server 2008 (MS). It looks like Gordon Linoff got the same error as I did. Also, the data type of NEWNUMBER is float. Given that I have floats, my ultimate goal is to count the number of decimal places to the right of the decimal.

Answer 1

You have got 3 for 222.9375 that's because REVERSE accepts a string expression. By passing in a float, SQL Server has to do an implicit conversion from float to varchar.

According to Books Online:

CAST and CONVERT (Transact-SQL)

float and real Styles
When expression is float or real, style can be one of the values shown in the following table. Other values are processed as 0.

Value
0 (default)

Output
A maximum of 6 digits. Use in scientific notation, when appropriate.

The above definition can be visualised using the examples below:

declare @a float = 222.9375, @b float = 12.34564, @c float = 1234564.123456
select convert(varchar, @a, 0) -- Returns 222.938 -- Notice only 6 digits are displayed
    , convert(varchar, @b, 0) -- Returns 12.3456 -- Notice only 6 digits are displayed
    , convert(varchar, @c, 0) -- Returns 1.23456e+006 -- Notice only 6 digits are displayed

If you keep using float as your data type, then you cannot expect to get the exact number you are looking for, even if you cast float into varchar and carry out string manipulation.

You cannot even simply cast float into decimal as you don't know how many decimal places (scale) to assign to the decimal.

Alternatively, if you start with decimal as the data type, then you automatically know how many decimal places a number has because you had to declare it.

For example

declare @number decimal(19, 4) = 222.9375
select @number -- 222.9375

Let's take it one step further. To work out the scale (number of decimal places) after some sort of calculation, you can employ the function below to get the answer:

declare @newnumber decimal(19, 4) = 222.9375    

select @newnumber * @newnumber -- Returns 49701.1289063
    , sql_variant_property(@newnumber * @newnumber, 'scale') -- Returns 7

However, you will most likely have no control over the data type used, one way I can think of is by combining the usage of both decimal and varchar to achieve what you want.

declare @newnumber float = 222.9375

-- The first 6 columns are just the working steps, solution is in the last column.
select 
    @newnumber as 'original'
    , cast(@newnumber as decimal(38, 18)) as 'decimal' -- Assuming all your numbers can fit in decimal(38, 18).
    , cast(cast(@newnumber as decimal(38, 18)) as varchar(max)) as 'varchar'
    , reverse(cast(cast(@newnumber as decimal(38, 18)) as varchar(max))) as 'reverse'
    , cast(reverse(cast(cast(@newnumber as decimal(38, 18)) as varchar(max))) as decimal(38, 0)) as 'decimal'
    , len(cast(reverse(cast(cast(@newnumber as decimal(38, 18)) as varchar(max))) as decimal(38, 0))) as 'len'

    , case charindex('.', @newnumber)
        when 0 then 0
        else len(cast(reverse(cast(cast(@newnumber as decimal(38, 18)) as varchar(max))) as decimal(38, 0)))
    end as 'All In One Step'