[El Forum]

[eluser][email protected][/eluser]
Is there any way i can round a decimal by just dropping off the last unecessary numbers wihtout rounding up or down???
ie.

3.459 to 3.45

or

5.999999 to 5.99

????

[El Forum]

[eluser]batteries[/eluser]
The php manual hint hint

.. but you don't wanna round, hmm. cast to string, and use string manipulation functions.

[El Forum]

[eluser][email protected][/eluser]
Thanks. That's what I was thinking...but maybe there was a simpler way I didn't know about.

[El Forum]

[eluser]linuxbz[/eluser]
How about this:

Code:

$x = floor($y*100)/100;

[El Forum]

[eluser]Nanodeath[/eluser]
I usually go with linuxbz's method. Just beware of the return types (round and floor both return a float for some reason).

[El Forum]

[eluser]linuxbz[/eluser]
[quote author="Nanodeath" date="1183458693"]Just beware of the return types (round and floor both return a float for some reason).[/quote]

Good point. Well, round() makes some sense because you can round it to a desired number of decimal places. The floor() function makes less sense, unless it is perhaps because you might want to floor() a larger number than the maximum integer size? I have heard that justification for a float before, though I can't think of an actual application of it.

Actually, as often as this kind of situation comes up, an argument to floor() like the number of decimals to round() might be handy for some people.

[El Forum]

[eluser]batteries[/eluser]
[quote author="linuxbz" date="1183458483"]

Code:

$x = floor($y*100)/100;

[/quote]

Nice code. Tiny and very clever.

[El Forum]

[eluser]linuxbz[/eluser]
@batteries - thanks.

Also, to display 3.30 rather than 3.3 try this:

Code:

$y = 5.9999;  // number to round down
$p = 2;  // number of decimals
$n = pow(10,$p);  // multiplier and divisor
$x = sprintf("#01.${p}f", floor($y*$n/$n));  // result is a string

which could be made into a function if needed more than once.

Edit: replace the # with % in the sprintf function above ... forum software wouldn't allow a % there apparently.