How do I modify this PHP code to return nested results

Question

I want to get this as a single object because I work with a JObject on the frontend.

I got an array at the moment but I am unsure how i should modify it so it returns a single object instead.

This is the code:

$contacts = array(); 

        while ($row = mysqli_fetch_array($stmt))
        {

            $contact = array("ID" => $row['ProduktID'],
                             "Name" => $row['ProduktNamn'],
                             "Number" => $row['ProduktPris']);

            array_push($contacts, $contact);
        }

        echo json_encode($contacts, JSON_PRETTY_PRINT);

And the goal is for it to look something like this with "results" as well so I can reach the whole thing:

What do you mean a single object? Don't loop and don't push onto the array if you don't want multiple elements in the array. — Devon Bessemer
– Devon Bessemer, Commented Feb 29, 2016 at 20:34

Viktor · Accepted Answer · 2016-02-29 20:42:19Z

2

To wrap your array of contacts in an object with a single results property:

echo json_encode(array('results' => $contacts), JSON_PRETTY_PRINT);

answered Feb 29, 2016 at 20:42

Viktor

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3 Comments

Devon Bessemer Over a year ago

I think this is what he actually wants, regardless of the confusing wording of his question. I was about to post it with a link to sandbox.onlinephpfunctions.com/code/…

William.John Over a year ago

I am sorry for being confusing! This did it though. Thanks alot guys. Amazing feedback and support. I am fairly new so this is amazing. Thx again.

William.John Over a year ago

Onto the next issue :) The life of a beginner programmer ;)

Sudoscience · Accepted Answer · 2016-02-29 20:31:44Z

2

You can use typecasting to convert an array to an object:

$object = (object) $array_name;

or manually convert it into an object

$object = object;
foreach($arr as $key => $value)
{
    $object->$key = $value;
}

answered Feb 29, 2016 at 20:31

Sudoscience

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2 Comments

William.John Over a year ago

Now I dont get 1 object I think. When I do it Each row gets "0" and the next one "1" etc.

William.John Over a year ago

Like this: { "0": { "ID": "43", "Name": "pro2", "Number": "pris2" }, "1": { "ID": "42", "Name": "Produkt1", "Number": "Pris1" } }

Xorifelse · Accepted Answer · 2016-02-29 20:51:44Z

0

Like this? Keep in mind an object is different then an array and your question is rather confusing.

    while ($row = mysqli_fetch_array($stmt)){
        $contact[] = [
          "ID" => $row['ProduktID'],
          "Name" => $row['ProduktNamn'],
          "Number" => $row['ProduktPris']
        ];
    }

    json_encode(['result' => $contact]); // Forget the JSON_PRETTY_PRINT.

Using this method [], it will use the first available numeric index starting with 0. This way you don't have to push an array.

edited Feb 29, 2016 at 20:51

answered Feb 29, 2016 at 20:36

Xorifelse

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5 Comments

William.John Over a year ago

I do not want it as an array. I want to send the info as a object. Just like the image I showed. Because I want to work with it as a JObject in my frontend

Devon Bessemer Over a year ago

@William.John, you're sending a JSON string, not an array. json_encode will encode it whether it's a PHP array or a PHP object. It makes no difference.

William.John Over a year ago

Hm okay. I was told otherwise in my previous thread (where I show my front end as well) You can take a look here if u want to see if there is something wrong in my code there: stackoverflow.com/questions/35705061/…

Xorifelse Over a year ago

@William.John As Devon said, your output is JSON its the array structure you need to worry about. And seeing the comment in Sudoscience answer this code would suffice.

William.John Over a year ago

I think Viktor just nailed it. I will try it and see

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