struggling to walk through php array element in recursive function

Question

I am using recursive function to walk through array elements.But failed to get correct counter.What i did.

$i = 0 ;
$this->make_items($items,$i);
public function make_items($items,$i)
{
    foreach($items as $item){
        $i++;
        echo  '<position>'.$i.'</position>';    
        if(count($item['children'])){
                $this->make_items($item['children'],$i);
        }
}

But for array structure like

arr1
arr2
    arr3
        arr4
   arr5
arr6

Output of my code is

<postion>1</position>
<postion>2</position>
<postion>3</position>
<postion>4</position>
<postion>4</position>
<postion>3</position>

While i needed

<postion>1</position>
<postion>2</position>
<postion>3</position>
<postion>4</position>
<postion>5</position>
<postion>6</position>

What i am doing wrong or what i understood wrong about recursion. If there is better way to do this then please advice. thanks in advance.

Jirka Picek · Accepted Answer · 2017-12-13 11:03:11Z

1

The variable $i changes only in the function and it doesn't affect variable outside of the function. When it's recursion, then it seems that it is in one function, but new variable $i is created for each call.

The simplest solution is pass value by reference. It's done by adding & before parameter $i in definition of function:

public function make_items($items,&$i) { /* the rest of code remains same*/ }

Second solution returning value of $i and assigning it as new value outside of called function:

$i = 0 ;
$this->make_items($items,$i); //now $i is still 0 but return value is 6

public function make_items($items,$i)
{
    foreach($items as $item){
        $i++;
        echo  '<position>'.$i.'</position>';    
        if(count($item['children'])){
                $i = $this->make_items($item['children'],$i); // assign new value of $i
        }
    return $i; // return actual value of $i
}

edited Dec 13, 2017 at 11:03

answered Dec 13, 2017 at 10:59

Jirka Picek

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2 Comments

Lawrence Cherone Over a year ago

Warning: Invalid argument supplied for foreach()

rahul singh Over a year ago

thanks for the short explanation .It worked I understood my mistake

G4Hu · Accepted Answer · 2017-12-13 10:36:50Z

0

Change your if (count($item)) to if (is_array($item))

answered Dec 13, 2017 at 10:36

G4Hu

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