How to json_decode invalid json?

Question

example.php

{"status": "ok"} {"status": "error"}

result:

ok , error

My website is showing a blank page(I'm using this code),can you help me to fix it?

<?php
$userinfo = 'example.php';
$fgc = file_get_contents($userinfo);
$json2 = json_decode($fgc, true);
$media = $json2['status'];

$mediaId = $media;

echo $mediaId;
?>

You may have to use the fully qualified path for example.php. E.g if `example.php` is in the root directory (ie `public_html`) then `$userinfo = $_SERVER['DOCUMENT_ROOT'] . '/example.php';` may resolve your issue — zanderwar, Dec 12 '16 at 03:20
You should enable error reporting: Add `ini_set('display_errors', 1); ini_set('error_reporting', E_ALL);` after ` — zanderwar, Dec 12 '16 at 03:20
In fact, my problem is in {"status": "ok"} {"status": "error"}, because there is no comma separating them — ALFIAN ANANDA PUTRA, Dec 12 '16 at 03:27
As robbie says, that's not valid JSON, either complain to the source or `str_replace("} {", "},{", $fgc)` with a risk at using the latter — zanderwar, Dec 12 '16 at 03:27
it's not safe, but if your JSON is literally as small as what you have provided, that str replace will give you valid JSON — zanderwar, Dec 12 '16 at 03:31
$fgc = file_get_contents($userinfo); $ber = str_replace("} {", "},{", $fgc); $json2 = json_decode($ber, true); ?? — ALFIAN ANANDA PUTRA, Dec 12 '16 at 03:35
yassss, i will result show "ok & error" ,can you give me the code? — ALFIAN ANANDA PUTRA, Dec 12 '16 at 03:40
There you go, see my answer - should get you back on the right track — zanderwar, Dec 12 '16 at 03:47

score 1 · Answer 1 · edited May 23 '17 at 12:08

1

example.php does not contain valid JSON

Here's probably what you want:

 [ {"status": "ok"} , {"status": "error"} ]

You can validate JOSN here: http://jsonlint.com/

You can find more on how to debug PHP "white screen of death" here: PHP's white screen of death (read the top two answers)

edited May 23 '17 at 12:08

Community

1
1

answered Dec 12 '16 at 03:26

Robbie

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4
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I took the results of external websites, and as a result there is no comma separating between it. – ALFIAN ANANDA PUTRA Dec 12 '16 at 03:30
Just because you got it from an external website does not mean it's right. It's NOT valid JSON, therefore `json_decode` WILL fail and throw an error and give you a white screen. – Robbie Dec 12 '16 at 03:56
1

@ALFIANANANDAPUTRA then contact whoever made this external website and tell him to learn the basics of what he's supposed to do, because that was **not** valid JSON. – Franz Gleichmann Dec 12 '16 at 06:59

score 0 · Accepted Answer · answered Dec 12 '16 at 03:45

This will resolve your issue in the non-recommended way:

$broken_json = '{"success": "ok"} {"success": "error"}';
$fixed_json = "[" . str_replace("} {", "},{", $broken_json) . "]";
echo $fixed_json;

$array = json_decode($fixed_json, true);

echo "<pre>";
var_dump($array);
echo "</pre>";

Result:

[{"success": "ok"},{"success": "error"}]

array(2) {
  [0]=>
  array(1) {
    ["success"]=>
    string(2) "ok"
  }
  [1]=>
  array(1) {
    ["success"]=>
    string(5) "error"
  }
}

Recommended Way:

Actually get VALID JSON from the source

How to json_decode invalid json?

2 Answers2