Passing of variables outside the loop in PHP - php

Question

I have a simple code that goes like this
<?php
echo $passer;
for($i=1; $i<=5; $i++){
$msg = $i;
}
$passer = $msg;
?>
My aim is to display the result above the loop. Is there a way to pass the value of $msg so it will be displayed above the loop? Currently the output is:
Undefined variable: passer

Answer 1

First you are getting error Undefined variable: passer. because you are using variable without defining. So define variable before using:
$passer = 0; //defining variable
echo $passer;// you are getting error here
for($i=1; $i<=5; $i++){
$msg = $i;
}
$passer = $msg;
Above code will not give you output as you expecting. because you are echoing $passer before initializing it. Try something like this:
for($i=1; $i<=5; $i++){
$msg = $i;
}
$passer = $msg;//initializing first
echo $passer;//output 5
Still in above solution i am echoing $passer after processing of for loop.
Because parser follows top to bottom approach so its impossible output
above loop.

Answer 2

$msg = array();
for($i=1; $i<=5; $i++){
$msg[] = $i;
}
$passer = $msg;
print_r($passer);

Answer 3

No One Language can output value before define that.
but you can like this to get your result.
<?php ob_start() ?>
##passer##
<?php
for($i=1; $i<=5; $i++){
$msg = $i;
} ?>
<?php echo str_replace("##passer##", $msg, ob_get_clean()) ?>