How to create this JSON in PHP?

Question

Sorry for the bad title, but I don't know how to create following JSON in PHP:

{ 
   "id":"1",
   "method":"getData",
   "params":{ 
      "options":{ 
         "element":{ 
            "id":"1a_ext",
            "type":1,
            "keyType":"externalkey"
         },
         "moreInfo":true,
         "userFields":[ 
            "id",
            "name",
            "longname",
            "externalkey"
         ]
      }
   },
   "jsonrpc":"2.0"
}

I don't know to do the part after "params" (how do I "put" options "into" params) - for the other parts I know what I have to do:

public static function getData(){
            $json = array(
                "id" => self::id(),
                "method" => "getData",
                "params" => array(
                    "id" => self::$userid,
                    "type" => self::$type
                ),
                "jsonrpc" => "2.0"
            );
            $json = json_encode($json, true);
            return self::request($json);
        }

I would really appreciate your help, thanks!

You can check result of json_decode($jsonString, true), for example. This will return array as you need and using this info you will create needed method — Dmytro
– Dmytro, Commented Sep 20, 2019 at 10:18

Adnan Ahsan · Accepted Answer · 2019-09-20 10:19:04Z

3

You directly can assign to the params keys like

$json['params']['options'] = $your_options;

Full version of your code as an example

public static function getData(){
    $json = array(
        "id" => self::id(),
        "method" => "getData",
        "params" => array(
            "id" => self::$userid,
            "type" => self::$type
        ),
        "jsonrpc" => "2.0"
    );

    # add something to param index
    $json['params']['options'] = $your_options;

    $json = json_encode($json, true);
    return self::request($json);
}

answered Sep 20, 2019 at 10:19

Adnan Ahsan

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3 Comments

Arnas Kazlauskas Over a year ago

Why are you passing true as a second param for json_encode?

Adnan Ahsan Over a year ago

for bitmasking during encoding, we have to pass true as an option.

Arnas Kazlauskas Over a year ago

The second param excepts an int (a flag) not a boolean value.

DrRoach · Accepted Answer · 2019-09-20 10:19:24Z

1

You can create the this in array format in PHP and then JSON encode:

$arr = [
    'id' => 1,
    'method' => 'getData',
    'params' => [
        'options' => [
            'element' => [
                'id' => '1a_ext',
                'type' => 1,
                'keyType' => 'externalKey'
            ],
            'moreInfo' => true,
            'userFields' => [
                'id',
                'name',
                'longname',
                'externalKey'
            ]
        ]
    ],
    'jsonrpc' => '2.0'
];

$json = json_encode($arr);

answered Sep 20, 2019 at 10:19

DrRoach

1,3569 silver badges16 bronze badges

Comments

lazyCoder · Accepted Answer · 2019-09-20 11:01:40Z

Instead of spoonfeeding, i would like to help related, Whenever if you find difficulties to create an array representation of a JSON then you should use var_export(array, true) the second parameter must be true to return the variable representation instead of outputting it

<?php
    $json_str = '{
      "id": "1",
      "method": "getData",
      "params": {
        "options": {
          "element": {"id": "1a_ext", "type": 1, "keyType": "externalkey"},
          "moreInfo": true,
          "userFields": ["id", "name", "longname", "externalkey"]
        }
      },
      "jsonrpc": "2.0"
    }';
    $json_arr = var_export(json_decode($json_str, true), true);
    print_r($json_arr);

check the output here https://paiza.io/projects/eUZZDsTsSFSM4m9WMl05Ow

$json_arr is an array representation for your JSON, now you can dynamic the array values

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