json_decode returns string type instead of object

Question

I\'m passing a JSON-encoded string to json_decode() and am expecting its output to be an object type, but am getting a string type instead. How can I return an

Answer 1

The function json_encode is used to encode a native PHP object or array in JSON format.

For example, $json = json_encode($arr) where $arr is

$arr = array(
  'a' => 1,
  'b' => 2,
  'c' => 3,
  'd' => 4,
  'e' => 5,
);

would return the string $json = '{"a": 1, "b": 2, "c": 3, "d": 4, "e": 5}'. At this point, you do not need to encode it again with json_encode!

To obtain your array back, simply do json_decode($json, true).

If you omit the true from the call to json_decode you'll obtain an instance of stdClass instead, with the various properties specified in the JSON string.

For more references, see:

http://www.php.net/manual/en/function.json-encode.php

http://www.php.net/manual/en/function.json-decode.php

Answer 2

Instead of writing on the JSON array, try putting it into a PHP array first.

<?php
$array = array(
'a' => 1,
'b' => 2,
'c' => 3,
'd' => 4,
'e' => 5
);
//Then json_encode()
$json = json_encode($array);
echo $json;
die;
?>

In you case, you are using ajax. So when you get a success, you can do this:

$.ajax({
    url: 'example.com',
    data: {

    },
    success: function(data) {
        console.log(data);
    }
});

Where after data inside console.log() can add the json var like data.a, data.b...

Also, with the string you providedm you do not need to json_encode since it is alrady in json format

Answer 3

var_dump(json_decode($json, true));

http://hk.php.net/manual/en/function.json-decode.php