If ... do this if statement? - php

Question

I am stuck on some php coding, that seemed initially easy to accomplish. Here is what i would like to do :
<?php
$amountOfDigits = 1;
$numbers = range(1,3);
shuffle($numbers);
for($i = 0;$i < $amountOfDigits;$i++)
$digits .= $numbers[$i];
while ( have_posts() ) : the_post();
static $count = 0;
if ($digits == '1') {
//Do this if statement
if ($count == "2") }
elseif ($digits == '2') {
//Do this if statement
if ($count == "2" || $count == "3") }
elseif ($digits == '3') {
//Do this if statement
if ($count == "2" || $count == "3" || $count == "4") }
{ //here the rest of the code
?>
So depending on the $digits variable, the if statement is formatted to be used on the line above //the rest of the code
How to put this in PHP properly?
Thanks.
Robbert

Answer 1

If I understand your question properly, you want something like this:
if ($digits == '1')
$mycond = ($count == "2");
elseif ($digits == '2')
$mycond = ($count == "2" || $count == "3")
elseif ($digits == '3')
$mycond = ($count == "2" || $count == "3" || $count == "4")
then you can further use
if($mycond){
// blahblahblah
}

Answer 2

Well if you want the same execution block on each case, there is a very simple solution for that.
You should simply use a function that check status of "count" depending on "digits".
<?php
function checkCountAgainstDigits($count, $digits){
switch($digits){
case 1:
return $count === 1;
case 2:
return $count === 2 || $count === 3;
case 3:
return $count === 2 || $count === 3 || $count === 4;
default:
// The default case "ELSE"
return FALSE;
}
}
if(checkCountAgainstDigits($count, $digits)){
// do
}
?>
If you want a different one, your solution is correct.