Insert an array into another array on same values

Question

I have this array:

ARRAY 1

Array
(
    [0] => Array
        (
            [required] => Tip:
        )    
    [1] => Array
        (
            [required] => Flux:
        )   
    [2] => Array
)

ARRAY 2

Array
(
    [0] => Array
        (
            [name] => Flux:
        )   
    [1] => Array
        (
            [name] => Tip:
        )   
    [2] => Array
        (
            [name] => Weight:
        )   
)

How insert data from array 1 to array 2 when value of name and required are same. Data from array some time don't have same values of name and required. On output of array 2 I want to display something when value of required exist.

I need this:

Array
(
    [0] => Array
        (
            [name] => Flux:
            [required] => Flux:
        )   
    [1] => Array
        (
            [name] => Tip:
            [required] => Tip:
        )    
    [2] => Array
        (
            [name] => Weight:
            [required] =>
        )    
)

I tried with this:

foreach($char as $key =>$value){
                foreach($mandatory as $key =>$val){     
                $data[] = array("name" => $value->charact_name, "required" => $val);                    
                }
            }

But is not output what I expect! It show me diferent values on name and required

Have you any tried? Please include your code in the question. — Amirreza Noori
– Amirreza Noori, Commented Dec 26, 2021 at 8:16

Amirreza Noori · Accepted Answer · 2021-12-26 08:36:39Z

1

This question is not a technical problem and you needed a little more tries to achieve a true algorithm. I did some modifications to your code. $key variable must be different for two foreachs. Based on your question new key required just added to $arr2.

$arr1 = [['required' => 'Tip:'], ['required' => 'Flux:']];
$arr2 = [['name' => 'Flux:'], ['name' => 'Tip:'], ['name' => 'Weight:']];

foreach($arr2 as $key2 => $value2) {
    $arr2[$key2]['required'] = NULL;
    foreach($arr1 as $value1) 
        if($value2['name'] == $value1['required']) { 
            $arr2[$key2]['required'] = $value1['required'];
            break;
        }
}

answered Dec 26, 2021 at 8:36

Amirreza Noori

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Comments

Fan Beko · Accepted Answer · 2021-12-26 09:03:07Z

0

I tried this:

foreach($char as $key2 => $value2) {
            foreach($mandatory as $value1)              
                if($value2->charact_name == $value1) { 
                    $data[] = array("name" => $value2->charact_name, "required" => $value1);    
                }
        }

But data array it show me name and required only for values that is in $mandatory array, I want to display all from $char and when value exist in $mandatory add [required]=> name of value from $mandatory

answered Dec 26, 2021 at 9:03

Fan Beko

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1 Comment

Fan Beko Over a year ago

Got it! I put $char->required = $value1; in if and data[] I put in first foreach and I got what I expected. Thanks @Amirreza Noori, from you I resolved this!

Majid Hajibaba · Accepted Answer · 2021-12-26 09:17:12Z

You can simply create it with nested loops which cause time complexity O(n*m). However if you use a Hashmap (array with key value) which get O(1) to search you can create it very fast by O(n+m) as below:

<?php
$arr1 = array(0 => array('name' => 'Flux'),1 => array('name' => 'Tip'),2 => array('name' => 'Weight'));

$arr2 = array(0 => array('required' => 'Tip'),1 => array('required' => 'Flux'),2 => array());


$arrkey = array();

foreach ($arr1 as  $ix=>$value) {
    $arrkey[$value['name']]=$ix;
}

$arr3 = $arr1;
foreach ($arr2 as $elem) {
    if(!empty($elem) && array_key_exists($elem['required'],$arrkey)) {
        $arr3[$arrkey[$elem['required']]]['required'] = $elem['required'];
        
    }
}


print_r($arr3);

?>

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