pushing string into empty array gives warning - php

Question

I created empty array and pushing val into it.
$var = array();
function printTag($tags) {
foreach($tags as $t) {
echo $t['token'] . "/" . $t['tag'] . " ";
if($t['tag'] == 'NN' OR $t['tag']== 'JJ'){
array_push($var, $t['token']) ;
}
}
echo "\n";
}
code looks fine for me but gives error:
array_push() expects parameter 1 to be array, null given in /var/www/html/postag/poscall.php on line 9
what is wrong here?
entire code:
<?php
// little helper function to print the results
include ("postag.php");
$var = array();
function printTag($tags) {
foreach($tags as $t) {
echo $t['token'] . "/" . $t['tag'] . " ";
if($t['tag'] == 'NN' OR $t['tag']== 'JJ'){
array_push($var, $t['token']) ;
}
}
echo "\n";
}
$tagger = new PosTagger('lexicon.txt');
$tags = $tagger->tag('The quick brown fox jumped over the lazy dog. this is really yummy and excellent pizza I have seen have really in love it it');
printTag($tags);
?>

Answer 1

Your $var = array(); statement is outside the function and out of the scope of the function. Put that inside the function and it'll remove the warning
function printTag($tags) {
$var = array();
foreach($tags as $t) {
echo $t['token'] . "/" . $t['tag'] . " ";
if($t['tag'] == 'NN' OR $t['tag']== 'JJ'){
array_push($var, $t['token']) ;
}
}
echo "\n";
}

Answer 2

The problem in your case is that $var is not in the scope of your function, so it gets implicitly declared as null (and this raises a notice too).
That said, this seems like a good case array_reduce():
$var = array_reduce($tags, function(&$result, $t) {
if (in_array($t['tag'], ['NN', 'JJ'])) {
$result[] = $t['token'];
}
// you could do some output here as well
return $result;
}, []);
It filters and maps at the same time and the return value is the array you want.
Alternatively, just declare $var inside the function and return it:
function printTag(array $tags)
{
$var = [];
foreach($tags as $t) {
// ...
}
return $var;
}
// ...
$var = printTag($tags);