Replace array value to increment

Question

I have an array like this ...

[0,0,23,0,0,18,0,0]

Then I want to change values that are not '0' (23 & 18) to auto increment, so the end result will be like this,

[0,0,1,0,0,2,0,0]

is there the best way for all that? So far this is what I did, but the results were not as expected ... :)

<?php

$arr = [0,0,23,0,0,18,0,0];
$x = 1;
$r = collect($arr)->map(function ($value, $key)use($x) {
    if ($value == 0) {
        return $value;
    } else {
        return $x++;
    }
})->all();

dd($r);

@dnFer I think it's probably [this](https://laravel.com/docs/5.7/helpers#method-collect). — Don't Panic, Nov 13 '18 at 21:28

score 3 · Accepted Answer · answered Nov 13 '18 at 20:43

The problem is that when you have your value of $x passed into the function via use, this isn't allowing the value to be updated. You need to pass it by reference - use (&$x) to allow it to increment the value outside the function...

$r = collect($arr)->map(function ($value, $key) use (&$x) {

Shahnawaz Kadari · Answer 2 · 2018-11-14T09:21:16.203

0

Another way... Using foreach with reference of array &

With Passed by Reference

Snippet

$arr = [0,0,23,0,0,18,0,0];
$counter = 1;
foreach ($arr as &$val){
   if($val !== 0){
      $val = $counter++;
   }
}
print_r($arr);

Note: Be aware of dangling reference

Without reference

Snippet

$arr = [0,0,23,0,0,18,0,0];
    $counter = 1;
    foreach ($arr as $key => $val){
       if($val !== 0){
          $arr[$key] = $counter++;
       }
    }
print_r($arr);

Output

Array
(
    [0] => 0
    [1] => 0
    [2] => 1
    [3] => 0
    [4] => 0
    [5] => 2
    [6] => 0
    [7] => 0
)

Live demo
Pass by reference docs

edited Nov 14 '18 at 09:21

answered Nov 13 '18 at 20:51

Shahnawaz Kadari

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`$val = $counter++;` or `$val = ++$counter;` saves you a line of code. ;) – DigiLive Nov 13 '18 at 21:13
Not when I change your demo. – DigiLive Nov 13 '18 at 21:30
@dnFer Link modified demo – Shahnawaz Kadari Nov 13 '18 at 21:33
@dnFer Thanks, modified. – Shahnawaz Kadari Nov 13 '18 at 21:41
@Progrock look at `foreach`, passed array by reference, that's why array is being altered directly. – Shahnawaz Kadari Nov 14 '18 at 08:57
@Progrock Since, OP was forgot to pass by reference, nigel gave correct answer, mine is just minified version. I've linked `Pass By Reference` docs, so OP and other mah learn more about it. – Shahnawaz Kadari Nov 14 '18 at 09:03
Safer: `$c = 1; foreach($data as $k => $v) if($v) $data[$k] = $c++;` Beware dangling references: https://3v4l.org/GqC1B – Progrock Nov 14 '18 at 09:08
@Progrock It's reference, Altering actual array without creating temporary array. – Shahnawaz Kadari Nov 14 '18 at 09:09
@Progrock I've added your code too! Thanks for letting me know about `Dangling Reference` – Shahnawaz Kadari Nov 14 '18 at 09:22

Replace array value to increment

2 Answers2