Counting items in an array having X attribute

Apparently, I can only count to 1

[riverstyx]

Hi, folks -

I could use some insight into solving what should be an easy task. Example:

I want to randomly select 5 cards from a standard playing card deck and count how many of each suit are returned:

[connect]
$sql = "SELECT * FROM cards ORDER BY RAND()LIMIT 5";
$result = mysql_query($sql);
while($cards = mysql_fetch_array($result)) {

$value = $cards["card_value"];
$suit = $cards["card_suit"];
$image = $cards["card_image"];
--------

Using count() within the loop returns a count of 1 since each card can only be of one suit. I tried using something like this:

$spades = 0;
if($suit='spades') {
$spades = $spades+1;
}

and I tried a foreach statement to count within the loop which always returns a count of 1.

Would someone be so kind as to point me in the right direction?

RS

[bomburmusicmallet]

Not sure if your code is cut-n-paste, but if it is, this is incorrect:

if($suit='spades') {
$spades = $spades+1;
}

It should be:

if($suit=='spades')

Notice the double equal sign.

[riverstyx]

Well, that did help a lot. I did this within the loop:

if($card_suit == 'spades') {
$spades=$spades+1;
}
if($card_suit == 'hearts') {
$hearts=$hearts+1;
}
if($card_suit == 'clubs') {
$clubs=$clubs+1;
}
if($card_suit == 'diamonds') {
$diamonds=$diamonds+1;
}
and then printed the counts outside the loop and everything is being counted correctly now. Thanks...

[IanKelley]

It's good to program efficiently right from the start.

That said, your loop will test every if statement even though only one of them can be true. Replace the last 3 if's with elseif or use continue instead.

[coopster]

True. A switch [php.net] control structure would work well too.

[whoisgregg]

So long as you can trust the $card_suit variable, you could use variable variables [php.net].

${$card_suit}++;