PHP json_decode error: Array value found, but an object is required

Question

I am trying to decode following JSON file

{
     "email": "mail@gmail.com",
     "password": "12345",
     "languageProficiency": {
            "language": "English",
             "proficiency": 4
   },
     "tags": [
         {
            "name": "singing"
         },
         {
           "name": "dance"
         }
   ]
}

When I do this

 $data = json_decode($jsonContent, true);
 echo $data;
 die();

I have following error:

Array value found, but an object is required

Question

1) How can I view data from JSON 2) How can I access property of each object in tags array

I am validating Json content againsts this schema

  {
    "$schema": "http://json-schema.org/draft-04/schema#",
    "type": "object",
    "properties": {
      "email": {
        "type": "string"
      },
      "password": {
        "type": "string"
      },
      "languageProficiency": {
        "type": "object",
        "properties": {
          "language": {
            "type": "string"
          },
          "proficiency": {
            "type": "integer"
          }
        },
        "required": [
          "language",
          "proficiency"
        ]
      },
      "tags": {
        "type": "array",
        "items": {
          "type": "object",
          "properties": {
            "name": {
              "type": "string"
            }
          },
          "required": [
            "name"
          ]
        }
      }
    },
    "required": [
      "email",
      "password",
      "languageProficiency",
      "tags"
    ]
  }

UPDATE

I tried following to view json content

 print_r($data)

But I still get same error

For this code you'll get "Notice: Array to string conversion". So your error comes from other place. — Gino Pane, Jul 28 '16 at 11:01

score 0 · Accepted Answer · answered Jul 28 '16 at 10:59

$data = json_decode($jsonContent, true);

Above line of code will return Array, for which you can not use echo directly to print an array.

To print specific value of an array (e.g. email), do it like this:

echo $data["email"];

Tip:

Use print_r() to know the array structure like this,

echo "<pre>";
print_r($data);

score 0 · Answer 2 · answered Jul 28 '16 at 11:00

0

you can try print_r($data) to explore values

answered Jul 28 '16 at 11:00

PHP json_decode error: Array value found, but an object is required

2 Answers2