PHP json_encode returning empty sctructure

Question

I'm calling json_encode to sendo a objects array back to the user, and at other parts of my code it works properly, but at this one it return a empty structure string.

Here is the result when I call json_encode on the array before encoding:

array(3) {
  [0]=>
  object(MinaProxy)#5 (7) {
    ["carregado":"MinaProxy":private]=>
    bool(true)
    ["link":"MinaProxy":private]=>
    int(1)
    ["idMina":protected]=>
    int(1)
    ["qntOuro":protected]=>
    int(2000)
    ["x":protected]=>
    int(307)
    ["idPartida":protected]=>
    int(1)
    ["proximo":protected]=>
    int(1)
  }
  [1]=>
  object(MinaProxy)#6 (7) {
    ["carregado":"MinaProxy":private]=>
    bool(true)
    ["link":"MinaProxy":private]=>
    int(2)
    ["idMina":protected]=>
    int(2)
    ["qntOuro":protected]=>
    int(2000)
    ["x":protected]=>
    int(512)
    ["idPartida":protected]=>
    int(1)
    ["proximo":protected]=>
    int(2)
  }
  [2]=>
  object(MinaProxy)#7 (7) {
    ["carregado":"MinaProxy":private]=>
    bool(true)
    ["link":"MinaProxy":private]=>
    int(3)
    ["idMina":protected]=>
    int(3)
    ["qntOuro":protected]=>
    int(2000)
    ["x":protected]=>
    int(716)
    ["idPartida":protected]=>
    int(1)
    ["proximo":protected]=>
    NULL
  }
}

Here is the result after json_encode:

[{},{},{}]

As can be noticed, there is no special characters in the original array, so I do not think it is a encoding issue. Here is the code where I call json_encode:

elseif($dados['acao'] == 'recuperarMinas') {
    $i = 0;
    $array = array();
    while ($this->partida->minas->temProximo()) {
        $array[$i] = $this->partida->minas->getProximoAvancando();
        $array[$i]->getIdPartida();
        $i++;
    }
    $_SESSION['partida'] = $this->partida;
    $retornoJson = json_encode($array);
    return $retornoJson;
}

score 5 · Answer 1 · edited Nov 22 '15 at 22:36

5

Problem is that your output has three Objects with private and protected attributes.

See this example

<?php

class sampleClass
{
    private $hello = 25;
}


$class = new sampleClass();

echo json_encode($class);

?>

Output will be {}

In your case you have array [] of three {} objects, and they are empty because all their attributes are private or protected.

But if you will change object attributes to be public, public $hello = 25; then output will be

{"hello":25}

edited Nov 22 '15 at 22:36

Kirk Beard

9,569
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answered Nov 22 '15 at 20:42

user991

1,339
2
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1

@Andreas my first answer in stackoverflow )) – user991 Nov 22 '15 at 20:47
(I meant you came first answering this particular question) – Andreas Nov 22 '15 at 20:49
@user777 just realized that you 've written method you meant properties/attributes isn't it? ;) – Andreas Nov 22 '15 at 22:17
After thinking about properties/attributes I landed on [this](http://stackoverflow.com/questions/258469/what-is-the-difference-between-attribute-and-property). Dear Lord ....!! SO is always fun! – Andreas Nov 22 '15 at 22:38
Coming from Java, this was a nightmare ! But totally logical ! – Nicolas Zozol Feb 19 '16 at 11:39

score 1 · Accepted Answer · answered Nov 22 '15 at 20:50

1

You have protected and private attributes, so you can not use any functions, including json_encode() to access directly to your object's property.

Use JSONSerialize

answered Nov 22 '15 at 20:50

Halayem Anis

7,654
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score 0 · Answer 3 · edited Oct 18 '22 at 23:35

It's because they are not public If you want json_encode private/protected variables you have to implement JSON serialization your self.

You could try considering jsonserializable (php 5.4+)with caution!!!!!!! Lets say that you want to serialize private/protected properties of your objects in your response to some http requests but in other cases you may not want these properties to be serialized or you may already run code(even third party) that assumes they are not

anyhow if you want to expose all your object's properties the code in your fuction could be something like this:

public function somethinghere() {
    return get_object_vars($this);
}

The full signature of the function was left out in purpose because you should decide what you want to do according to your context(I was having considerations before posting this but I hope this way criticism will be minimized ) This example may also be considered as bad because it exposes everything(you may consider creating an associative array that contain only the properties you want to expose )

working example

PHP json_encode returning empty sctructure

3 Answers3

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