[Resolved] Know if string has at least 3 letters and 3 digits

[spighita]

Hello:
I would like to know how can I count the amount of digits and letters in a string.
The purpose is to validate a string.

I'm using ereg but my code fails when I use for example a1a2a3.

Thanks for anyone who helps.

[vallette]

You could just use a little routine like so:

$string='a1a2a3b';
$digits=0;
$characters=0;
$len=strlen($string);
for($i=0;$i<$len;$i++)
	{
	$value=ord($string[$i]);
	if($value>47&&$value<58)
		$digits++;
	else
		$characters++;
	}
echo "$digits $characters";

[LordShryku]

Hrmmm.....sure, but what if the string was
$string='a1a2a3)~';

How about instead...

$string     = 'a1a2a3)~';
$digits     = 0;
$characters = 0;
for($i=0;$i<strlen($string);$i++) {
   if(ctype_digit($string{$i})) { $digits++; }
   if(ctype_alpha($string{$i})) { $characters++; }
}

[spighita]

Thanks, both ideas are good.

I liked them both, but I'll use LordShryku's code.

I think it works faster.

Thank you very much

[Weedpacket]

Taking the title of the thread into account, a wee optimisation to LordShryku's code could be made by adding a break.

$string     = 'a1a2a3)~';
$digits     = 0;
$characters = 0;
for($i=0;$i<strlen($string);$i++) {
   if(ctype_digit($string{$i})) { $digits++; }
   if(ctype_alpha($string{$i})) { $characters++; }
  if($digits>=3 && $characters>=3) break;
}

If you've found at least three digits and three letters, you don't need to keep looking! It's probably not much of a saving unless the line is potentially a fair bit longer than six characters.

(And - just because there's more than one way to do it, not because it's necessarily a good idea - with regexps!)

$count_digits = preg_match_all('/[0-9]/',$string);
$count_letters = preg_match_all('/[a-z]/i', $string);