PHP: How do you convert from EXPONENT_DNUM to LNUM format?? - GoFuckYourself.com

Varius · 04-20-2005, 11:26 AM

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..

04-20-2005, 11:26 AM	#1
Varius Confirmed User Industry Role: Join Date: Jun 2004 Location: New York, NY Posts: 6,890	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. __________________ Skype variuscr - Email varius AT gmail

04-20-2005, 11:31 AM	#2
Varius Confirmed User Industry Role: Join Date: Jun 2004 Location: New York, NY Posts: 6,890	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]);` __________________ Skype variuscr - Email varius AT gmail

04-20-2005, 11:39 AM	#5
swedguy Confirmed User Industry Role: Join Date: Jan 2002 Posts: 7,981	err...... Last edited by swedguy; 04-20-2005 at 11:40 AM..

04-20-2005, 12:04 PM	#9
mortenb Confirmed User Join Date: Jul 2004 Location: Denmark ICQ: 7880009 Posts: 2,203	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..

04-20-2005, 11:34 AM	#3
pstation Confirmed User Join Date: Jul 2003 Location: chicago Posts: 1,135	use the force

04-20-2005, 11:36 AM	#4
azguy Confirmed User Join Date: Nov 2004 Location: Scottsdale, AZ Posts: 5,167	Try to add 1 to it

04-20-2005, 11:43 AM	#8
jeffs Confirmed User Join Date: Oct 2002 Location: chair Posts: 135	You can use printf("%f", $var) or printf("%.5f", $var) instead of echo $var.