Array replace not working

Question

The below is my array, where I need to replace the value of 'battle_health'

$battlepokemon= array();

$i = 1;
while($rows = mysql_fetch_assoc($res))
{
$path = mysql_query(" SELECT * FROM pokemons WHERE pk_id = '".$rows['pkmn_id']."' ");
$pokemon = array(
'opponent_increment' => $i,
'id' => $rows['pkmn_id'],
'battle_poke'=> mysql_result($path,0,"path"),
'battle_level' => $rows['level'],
'battle_health' => $rows['health']
);
$i++;
$battlepokemon[]= $pokemon;
}

The code for replacement is:

$i = 1;
foreach ($battlepokemon as $key => $value) 
{
if($value['opponent_increment'] == $opponent_increment)
{
$value['battle_health'] = 0;
echo "Data replaced!";
}
$i++;
}
print_r($battlepokemon);

The code above is working..from start to end.. but the value is not replaced with '0' as the code says! I think I must have missed something!

your replacing in the temporary copy of the array php creates for the loop, not the original — user557846
– user557846, Commented Aug 22, 2014 at 3:59
change foreach ($battlepokemon as $key => $value) to foreach ($battlepokemon as $key => &$value) — Iłya Bursov
– Iłya Bursov, Commented Aug 22, 2014 at 4:07

Bhavik Shah · Accepted Answer · 2014-08-22 04:38:34Z

4

You need to transfer the reference, not the values. Add a & to the following sentence

foreach ($battlepokemon as $key => &$value) 
                                   ^

I tried this just for example

<?php
$arr = array('12', '34');
foreach($arr as $key => &$value){
  $value = 0;
}
var_dump($arr);
?>

Hopes it can help you

edited Aug 22, 2014 at 4:38

Bhavik Shah

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answered Aug 22, 2014 at 4:10

iatboy

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Comments

Vivek Pratap Singh · Accepted Answer · 2014-08-22 05:06:05Z

0

You can achieve this with for Loop Because unlike foreach loop, it doesn't perform an array copy before transversal:

$arr = array('12', '34');
for($i = 0, $count = count($arr); $i < $count; $i++){
    $arr[$i] = 0;
}
var_dump($arr);

Or If you want to do with Foreach only, you need to avoid new copy of array by passing the reference like:

   $arr = array('12', '34');
   foreach($arr as $key => &$value)
   {
       $value = 0;
   }
   var_dump($arr);

edited Aug 22, 2014 at 5:06

answered Aug 22, 2014 at 4:21

Vivek Pratap Singh

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1 Comment

Scopey Over a year ago

Downvoted as the code you supplied is quite inefficient. You can use a foreach loop, you just need force the value to be a reference (noted in another answer) or refer to the scoped array variable. You must also be careful with a count in your for loop. The second argument of your for loop will be executed every iteration. The more efficient code is: for($i = 0, $count = count($arr); $i < $count; $i++)

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