PHP array loop by link and same name

Question

I can not understand. Why this code remove last item. But if i change $item in second loop on $item2, or use link, it work fine.

<?php
$list = [
    ['id' => 1],
    ['id' => 2],
    ['id' => 3],
    ['id' => 4],
    ['id' => 5],
    ['id' => 6],
];

$selected = [2,3,4,6];
$hidden = [4,5];

foreach ($list as &$item) {
    if(in_array($item['id'], $selected)) {
        $item['selected'] = true;
    }
}

foreach ($list as $key=>$item) {
    if(in_array($item['id'], $hidden)) {
        unset($list[$key]);
    }
}

var_dump($list);

use unset($item) after the first foreach to break the association of $item with the last array element — ewcz
– ewcz, Commented Mar 14, 2017 at 15:21

ewcz · Accepted Answer · 2017-03-14 15:32:18Z

2

as the documentation states:

Warning: Reference of a $value and the last array element remain even after the foreach loop. It is recommended to destroy it by unset().

so something like this should work:

foreach ($list as &$item) {
    //do some stuff
}
unset($item);

edited Mar 14, 2017 at 15:32

answered Mar 14, 2017 at 15:25

ewcz

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Comments

Luke Bradley · Accepted Answer · 2017-03-14 15:34:57Z

0

unset() the hidded keys first and then check for selected id's

foreach ($list as $key => $item) {
    if(in_array($item['id'],$hidden)){
        unset($list[$key]);
    } 
}

foreach ($list as &$item) {
    if(in_array($item['id'], $selected)) {
        $item['selected'] = true;
    }
}

edited Mar 14, 2017 at 15:34

answered Mar 14, 2017 at 15:29

Luke Bradley

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