Notice: Use of undefined constant for already defined variable [duplicate] - php

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Reference - What does this error mean in PHP?
(38 answers)
Closed 5 years ago.
I've got this code.
$rollcount=0;
$rollcounts=array(); //I define $rollcounts here
$number_of_tries = 100;
foreach(range(0,$number_of_tries-1) as $i){
do{
$roll=rand(1,6);
$rollcount++;
}while($roll!=6);
array_push($rollcounts, $rollcount);
$rollcount = 0;
}
$freqs = array();
while (!empty($rollcounts)){
$freq = count(array_filter($rollcounts,function($a) use ($rollcounts)
{return $a == $rollcounts[0];}
));
$freqs[$rollcounts[0]] = $freq;
for($i=0;$i<count($rollcounts);$i++){
if(rollcounts[$i] == $rollcounts[0]){ // THIS IS LINE 40
unset($rollcounts[$i]);
}
}
} // redo until $rollcounts is empty
That generates this error message (line 40 has been commented in the code)
Notice: Use of undefined constant rollcounts - assumed 'rollcounts'
in /Applications/XAMPP/xamppfiles/htdocs/learningphp/myfirstfile.php
on line 40
Clearly, $rollcounts has already been defined in the code. So what is the problem here?

Answer 1

You forgot a $
Old code
if(rollcounts[$i] == $rollcounts[0]){ // THIS IS LINE 40
unset($rollcounts[$i]);
}
New code
if($rollcounts[$i] == $rollcounts[0]){ // THIS IS LINE 40
unset($rollcounts[$i]);
}