The code is pretty straight forward, however it is just not running the way it is expected to, basically the numbers get added at the end in order to continue the while loop. But they are just not adding, the $two and $i that is
PHP:
$i=1;
$two=2;
$add=9;
$count=1;
while($i<=7488){
echo $count;
echo $exploded[$two];
echo "<br>count=".$count."<br>";
echo "<br>two=".$two."<br>";
echo "<br>i=".$i."<br>";
$count++;
$two+$add;
$i+$add;
}
How about this:
$two += $add;
$i += $add;
$two + $add is just an expression that returns the sum of the two variables; it doesn't actually do anything or change any state.
$two += $add (+= is the addition assignment operator) is equivalent to $two = $two + $add. Analogous operators exist for other arithmetic operations (e.g. *=, -=, etc.).
This pattern is true for all C-like languages (to my knowledge), and many other languages too.
+= resets the variable as the result?It doesn't work like that. $two+$add has a return value. You need to assign it to something.
You have to assign $two+$add;, $i+$add; to some var.
You do not modify this variables:
$two+$add;
$i+$add;
Maybe you need:
$two+=$add;
$i+=$add;
You're not saving your addition anywherE:
$two+$add;
$i+$add;
so the result of addition is simply tossed away.
Try:
$two = $two + $add;
or
$two += $add;
The term
$i+$add;
produces a result, but you throw that result away. If you would to save the result in $i again, you should try
$i = $i + $add;
or
$i += $add;
(Both statements do the same, the latter is just a shortcut.) The same holds for the other variables.
