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Posted by The Natural Philosopher on 02/01/08 16:40
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 $xxx is...
Oh my god.I AM GOING MAD..
I put in this one
printf ("%s<br>\r\n",$xxx);
AND THAT GIVES 6999 !!!!
WTF is php playing at?
I tried printf("%d" ,6999) and printf ("%d",'6999') and those work
But printf("%d",69.99*100) does NOT!
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