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Start your free trialBrane Opačič
6,686 PointsFunctions question in "Making a Basic PHP website"
Hello, good people.
I just can't grasp the logic here - I think the concatenation is giving me trouble. Why is the answer 321, can someone walk me through?
Thank you so much!
Code:
<?php
$numbers = array(1,2,3,4);
$total = count($numbers);
$sum = 0;
$output = "";
$i = 0;
foreach($numbers as $number) {
$i = $i + 1;
if ($i < $total) {
$output = $number . $output;
}
}
echo $output;
?>
2 Answers
Zachary Billingsley
6,187 PointsHello!
I'll try to break it down as best I can.
Here are your starting variables -> /********* $numbers = [1,2,3,4]; (This is another way to write an array) $total = 4; (count of the $numbers array) $sum = 0; (Doesn't seem to be used anywhere in the code though)
$output = ""; $i = 0; *********/
(1st loop) Inside your foreach loop, you start with $i = 1 ( $i = 0 + 1) and $number = 1 (the first index of your $numbers array). if ( 1 < 4 ) Which is TRUE $output = 1; (because the concatenation of 1 and "" is just 1)
(2nd loop) Now increment $i by 1 to get 2 and your $number = 2 (the second index of your $numbers array). if ( 2 < 4 ) Which is TRUE $output = 21; (because the concatenation of 2 and 1 is 21)
(3rd loop) Now increment $i by 1 to get 3 and your $number = 3(the third index of your $numbers array). if ( 3 < 4 ) Which is TRUE $output = 321; (because the concatenation of 3 and 21 is 321)
(4th loop) Now increment $i by 1 to get 4 and your $number = 4 (the fourth index of your $numbers array). if ( 4 < 4 ) Which is FALSE $output stays the same
echo $output; //Gives you 321
Hope that helps!
Brane Opačič
6,686 PointsWow! Thanks, Zachary! I needed the push you just gave me to continue working hard. Thank you so much.
Zachary Billingsley
6,187 PointsYou are very welcome!