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Posted by Jerry Stuckle on 02/01/08 16:37
The Natural Philosopher wrote:
> Rik Wasmus wrote:
>> On Fri, 01 Feb 2008 16:59:18 +0100, The Natural Philosopher <a@b.c>
>> wrote:
>>> The question says it all.
>>>
>>> I have an input box, which I fill in with a price.
>>> IOt gets passed to the main form as a variable, then shoved into an
>>> SQL field vue a print '%d' statement where the argument is $price*100.
>>
>> Could you give us the exact code & input values? And what database
>> (&version) are you using?
>>
>>> For some reaosn, this particular value goes to 6998.
>>>
>>> If I update hee database manually to 6999, it displays as 69.99.
>>>
>>> If I enter 69.999 it updates as 6999..not 69999
>>
>> Seems to have something to do with float/double precision. However,
>> the precision isn't that big, and I cannot reproduce it in PHP 5.2 here..
>>
>> PHP5.2.4:
>> <?php echo 69.99*100 ?> => output 6999
>> MySQL5.0.45-community-nt:
>> SELECT CAST(69.99*100 AS UNSIGNED); => 6999
>
> Rik. Its worse than that NOTHING to do with databases.
>
> Here is a code fragment.
>
> echo $sale_price."<br>\r\n"; // gives 69.99
> printf ("%d<br>\r\n",($sale_price*100)); //gives 6998
> $xxx=$sale_price*100;
> echo $xxx."<br>\r\n"; // gives 6999
> printf ("%d<br>\r\n",$xxx); //gives 6998
>
>
> Now $sale_price comes from a text type input via a _POST_ variable, so I
> guess its 'text' as its raw form PHP invisible casting drives me nuts,
> so you can tell ME what $xxx is...
>
According to the PHP manual:
%d - the argument is treated as an integer, and presented as a (signed)
decimal number.
Which means it is truncating instead of rounding.
69.99 is not exact. It's approximately 69.98999999999999488409.
So 69.99*100 comes out to be 6998.999999999999488409 and truncated you
get 6998.
--
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Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex@attglobal.net
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