Filter and update an array with mixed keys using another array with mixed keys

Question

I have two arrays.

$arr1 = array(
    'name',
    'date' => array('default' => '2009-06-13', 'format' => 'short'),
    'address',
    'zipcode' => array('default' => 12345, 'hidden' => true)
);

$arr2 = array(
    'name',
    'language',
    'date' => array('format' => 'long', 'hidden' => true),
    'zipcode' => array('hidden' => false)
);

Here's the desired result:

$final = array(
    'name',
    'date' => array('default' => '2009-06-13', 'format' => 'long', 'hidden' => true),
    'zipcode' => array('default' => 12345, 'hidden' => false)
);

Only the elements from $arr2 (that also exist in $arr1) are used
Each element's attributes are merged
If a common element (e.g. zipcode) shares an attribute (e.g. hidden), then the attribute from $arr2 takes precedence

What are some good approaches for solving this problem?

I tried to hobble something together... critiques welcomed:**

$new_array = array_intersect_key($arr2, $arr1);

foreach ($new_array as $key => $val)
{
    if (is_array($arr1[$key]))
    {
        if (is_array($val))
        {
            $new_array[$key] = array_merge($val, $arr1[$key]);
        }
        else
        {
            $new_array[$key] = $arr1[$key];
        }
    }
}

Good practice in this case is to post code like that as an answer to your own question. — Eric
– Eric, Commented Jun 14, 2009 at 3:01
I was looking at this but am wondering about a slight inconsistency. You only care about matching for the first-dimension of the array, and then merge everything in the second dimension, overwriting? Also, will this ever extend past two dimensions? — Ian Elliott
– Ian Elliott, Commented Jun 14, 2009 at 3:12
This will never extend past two dimensions. And yes, the $arr1's element's attributes will get overwritten by any matching $arr2 attributes. — Matt
– Matt, Commented Jun 14, 2009 at 3:22

Ian Elliott · Accepted Answer · 2009-06-14 04:06:18Z

2

You were close

$newArr = array_intersect_key($arr1, $arr2);
foreach ($newArr as $key => $val)
{
    if (is_array($val))
    {
        $newArr[$key] = array_merge($arr1[$key], $arr2[$key]);
    }
}

Edit Just had to change the array_intersect to array_intersect_key

edited Jun 14, 2009 at 4:06

answered Jun 14, 2009 at 3:35

Ian Elliott

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2 Comments

Matt Over a year ago

That doesn't look quite right either. If $arr1['address'] is an array, then it still makes it to the final output... :\

Ian Elliott Over a year ago

I didn't really check it throughly, it just gave me the proper result so I stopped :x. Let me have another look.

mickmackusa · Accepted Answer · 2024-11-10 05:17:10Z

Your described logic appears to:

retain the indexed values of the first array which are found as indexed values of the second array, and
overwrite the subarray elements of associative values in the first array which with corresponding subarrays of the second array.

It is unclear/unknown if an associative element might not have a subarray, so this is not built into the sceipt below. Demo

$arr2List = array_filter(
    $arr2,
    is_int(...),
    ARRAY_FILTER_USE_KEY
);

$result = [];
foreach ($arr1 as $k => $v) {
    if (is_int($k)) {
        if (in_array($v, $arr2List)) {
            $result[] = $v;
        }
    } elseif (
        is_array($v)
        && is_array($arr2[$k] ?? null)
    ) {
        $result[$k] = array_replace($v, $arr2[$k]);
    }
}
var_export($result);

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