[RESOLVED] php coding for some mathematical purpose...

[chandan_tiwari]

hi,

I'm beginner to php.

Can any one tell me the logic used for the following :

suppose the user has entered his date of birth as
dob=28
mob = 12
yob = 1985
i wanted my php code to add these numbers as:
2+8=10=1+0=1(till single digit)
similarly,
1+2=3
& 1+9+8+5=23=2+3=5

and then it should add all the three no.
1+3+5=9.
the output should be the number 9 or whatever comes in this way.

I have made the database table for the dob, mob, yob differently.

please let me know the logic fo this....(or code)

Thanking you!

[scross]

Here's a function that can do the job for you:

function math_calc($num){
	$str = (string) $num;
	while(strlen($str) > 1){
		$num = $str[0];
		for($i=1; $i<strlen($str); $i++){
			$num += $str[$i];
		}
		$str = (string) $num;
	}
	return (int) $str;
}

Then just go:

$dob = math_calc($dob);
$mob = math_calc($mob);
$yob = math_calc($yob);
echo $dob + $mob + $yob;

That should do your interesting calculation for you 🙂

[laserlight]

Actually, repeatedly adding digits until you have a single digit is almost equivalent to taking the remainder of the number after dividing by 9. The difference is that if the input is non-zero, then a zero remainder should be taken as 9. Since you do not expect more than 4 digits for the year of birth, you are assured of no overflow, and that the input will not be zero, and thus can write:

function checkDigit($num) {
    $num %= 9;
    return ($num == 0) ? 9 : $num;
}

$dob_check = checkDigit($dob);
$mob_check = checkDigit($mob);
$yob_check = checkDigit($yob);
echo checkDigit($dob_check + $mob_check + $yob_check);

[scross]

Nice, that looks like a great (and fast) way to do it 😛

[chandan_tiwari]

Great,
Thanks a lot for all of you for giving me such useful concept and code for my logic.
I am now understanding the application of php quite well.

The forum seems to have very good for the php beginners, it gives us all the code and logic related to particular topic.

Thank you once again.