Remove level in a multidimensional array by replacing parent element with its subarray data

Question

I've got an multidimensional array.

{
    "id": "13",
    "name": "Example",
    "location_name": "NY",
    "phone": [
        {
            "number": "0617357707"
        }
    ],
    "facilities": {
        "data": [
            {
                "name": "AC"
            },
            {
                "name": "Wi-Fi"
            }
        ]
    }
}

The problem I want to solve is at "facilities". Inside facilities, there is a data object. How can I remove the data without removing the content "data"?

Expected result:

{
    "id": "13",
    "name": "Example",
    "location_name": "NY",
    "phone": [
        {
            "number": "0617357707"
        }
    ],
    "facilities": [
        {
            "name": "AC"
        },
        {
            "name": "Wi-Fi"
        }
    ]
}

I've tried using array_walk or flatten but not getting it right.

Your json representation is wrong. It's neither an object nor an array. Is only the facilities relevant to your question? Is the results parts of an object or an array? — k0pernikus
– k0pernikus, Commented Oct 28, 2015 at 15:04
oh yes thanks for pointing. i copy from my fractal result so it got "results". @k0pernikus hi i've update with my answer. is it correct? yes i got my expected result. but should i keep doing that everytime i want to remove "data"? — ssuhat
– ssuhat, Commented Oct 28, 2015 at 15:06
Don't add answers within the question. And shouldn't you know if an answer works for you? ;) Your JSON is still broken. Check it here before posting: jsonlint.com — k0pernikus
– k0pernikus, Commented Oct 28, 2015 at 15:10

Andrius · Accepted Answer · 2015-10-28 14:21:50Z

3

Looks like you could just do:

$results->facilities = $results->facilities->data

Working example:

<?php

$var = json_decode('{
    "id": "13",
    "name": "Example",
    "location_name": "NY",
    "phone": [
        { "number": "0617357707" }
    ],
    "facilities": {
        "data": [
            { "name": "AC" }, { "name": "Wi-Fi" }
        ]
    }
  }');
var_dump($var);

$var->facilities = $var->facilities->data;
var_dump($var);

edited Oct 28, 2015 at 14:21

answered Oct 28, 2015 at 14:12

Andrius

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4 Comments

ssuhat Over a year ago

not working. "Trying to get property of non-object". and if i do this, i won't get the other result right?

Andrius Over a year ago

I've added a working example. Obviously, you should first check if 'facilities' and 'data' exists before accessing that data. From the error you're getting, I'd say that 'facilities' does not exist in one of your results.

ssuhat Over a year ago

Hi. i managed to answer slightly different from yours. $result['facilities'] = $result['facilities']['data']; i don't know why i keep getting "trying to get propery of non object" using yours

innocenat Over a year ago

This answer is expecting an object while your actual data structure is an array. json_decode will turn JSON object into PHP object, while json_encode encode both associative array and object into JSON object.

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