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Varius

04-20-2005 11:26 AM

PHP: How do you convert from EXPONENT_DNUM to LNUM format??

I'm having a very annoying problem this morning....

If I increment a variable by small floats (like 0.00001 lets say) in a loop, the vairable becomes EXPONENT_DNUM format. I would like to keep it in LNUM format. Any ideas?

I've made a simple loop here to show an example:

PHP Code:


		
			
<?PHP



$var = 0;



for ($i=0;$i<5;$i++) {

    $var += 0.000004;

}



echo $var;



?>

That gives me "2E-05" instead of the 0.00002 I"m looking for.

Varius

04-20-2005 11:31 AM

I do have one way to make it work, but I'd prefer finding out the true method...my way is just a workaround :)

I replace the last line by this instead, and then it works:

PHP Code:


		
			
$num = split("-",$var);



echo number_format($var,$num[1]);

pstation

04-20-2005 11:34 AM

use the force

azguy

04-20-2005 11:36 AM

Try to add 1 to it

swedguy

04-20-2005 11:39 AM

err......

Varius

04-20-2005 11:41 AM

Quote:

Originally Posted by azguy

Try to add 1 to it

No luck...also tried * 1, using settype to make sure they are equivalent types and few other things.

The weird part is, if I only add a value once, its fine.

ie.

PHP Code:


		
			
$var = 0;

$var += 0.00004;



echo $var;

will give me 0.00004 as output.

Once I loop on it though and add more values, it changes format.

Weird, no one on IRC or google could solve it...my fix above though works for my purposes..

azguy

04-20-2005 11:42 AM

Quote:

Originally Posted by Varius

No luck...also tried * 1, using settype to make sure they are equivalent types and few other things.

The weird part is, if I only add a value once, its fine.

ie.

PHP Code:


		
			
$var = 0;

$var += 0.00004;



echo $var;

will give me 0.00004 as output.

Once I loop on it though and add more values, it changes format.

Weird, no one on IRC or google could solve it...my fix above though works for my purposes..

I'd stick to what works then :)

jeffs

04-20-2005 11:43 AM

You can use printf("%f", $var) or printf("%.5f", $var) instead of echo $var.

mortenb

04-20-2005 12:04 PM

PHP Code:


		
			


<?PHP

bcscale(6);

$var    = 0;



for ($i=0;$i<5;$i++) {

    $var    = bcadd($var,1.000004);

}

$var    = bcsub($var,5);



echo $var;

?>

dunno if that is close enough for you..

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