How to push array values via php function? [duplicate]

Question

I have an empty array that I am trying to push a value to via a simple php function. The problem is that within each iteration the values are not being retained. Here is an example:

function addColors($arrayValues, $arrayToUpdate){
    $arrayToUpdate[]=$arrayValues;
}

$colors = array();
$newColors= array("red", "blue", "yellow");

foreach($newColors as $newColor){
    addColors($newColor, $colors);
}

echo "<pre>".print_r($colors, true)."</pre>";

This will just print an empty array. Whereas what I would like to see are the values being added to the $colors array. Any suggestions?

Add & to $arrayToUpdate and $newColor in function definitions to keep it as reference. — Peon
– Peon, Commented Jun 13, 2017 at 15:35
Not quite sure what you mean, but perhaps you just need to pass the array by reference? function addColors($arrayValues, &$arrayToUpdate){ — Plasm
– Plasm, Commented Jun 13, 2017 at 15:35

AbraCadaver · Accepted Answer · 2017-06-13 15:53:54Z

2

You either need to return the new array and assign the returned array in the loop:

function addColors($arrayValues, $arrayToUpdate){
    $arrayToUpdate[]=$arrayValues;
    return $arrayToUpdate;
}

foreach($newColors as $newColor){
    $colors = addColors($newColor, $colors);
}

Or to do it the way you have it, pass the variable that needs to be updated as a reference; notice the &. This is my recommendation:

function addColors($arrayValues, &$arrayToUpdate){
    $arrayToUpdate[]=$arrayValues;
}

foreach($newColors as $newColor){
    addColors($newColor, $colors);
}

Though in this simple example I wouldn't use a function:

foreach($newColors as $newColor){
    $colors[] = $newColor;
}

Also, there is already a function that does this, though the arguments are in a different order:

array_push($colors, $newColor);

Even simpler without the loop:

$colors = array_merge($colors, $newColors);

edited Jun 13, 2017 at 15:53

answered Jun 13, 2017 at 15:35

AbraCadaver

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