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);

What have you tried so far?

Nigel Ren
– Nigel Ren

2018-11-13 20:27:42 +00:00
Commented Nov 13, 2018 at 20:27 — Nigel Ren
– Nigel Ren, Commented Nov 13, 2018 at 20:27
Show us some code!

Brogan
– Brogan

2018-11-13 20:30:23 +00:00
Commented Nov 13, 2018 at 20:30 — Brogan
– Brogan, Commented Nov 13, 2018 at 20:30
If you increment 23, does that not become 24?

Progrock
– Progrock

2018-11-13 20:32:16 +00:00
Commented Nov 13, 2018 at 20:32 — Progrock
– Progrock, Commented Nov 13, 2018 at 20:32
Out of curiosity... what's function collect?

DigiLive
– DigiLive

2018-11-13 21:08:07 +00:00
Commented Nov 13, 2018 at 21:08 — DigiLive
– DigiLive, Commented Nov 13, 2018 at 21:08
@dnFer I think it's probably this.

Don't Panic
– Don't Panic

2018-11-13 21:28:10 +00:00
Commented Nov 13, 2018 at 21:28 — Don't Panic
– Don't Panic, Commented Nov 13, 2018 at 21:28

Nigel Ren · Accepted Answer · 2018-11-13 20:43:28Z

3

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) {

answered Nov 13, 2018 at 20:43

Nigel Ren

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Comments

Shahnawaz Kadari · Accepted Answer · 2018-11-14 09:21:16Z

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, 2018 at 9:21

answered Nov 13, 2018 at 20:51

Shahnawaz Kadari

1,5911 gold badge14 silver badges21 bronze badges

9 Comments

DigiLive Over a year ago

$val = $counter++; or $val = ++$counter; saves you a line of code. ;)

DigiLive Over a year ago

Not when I change your demo.

Shahnawaz Kadari Over a year ago

@dnFer Link modified demo

Shahnawaz Kadari Over a year ago

@dnFer Thanks, modified.

Shahnawaz Kadari Over a year ago

@Progrock look at foreach, passed array by reference, that's why array is being altered directly.

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