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Posted by The Natural Philosopher on 02/01/08 16:29
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...
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