Can't decode JSON value in PHP

Question

I found no solution for this.

I have an API witch give me JSON resposne,

http://api.vajehyab.com/v2/public/?q=%D8%B3%D9%84%D8%A7%D9%85&developer=stackoverflow

which is something like this,

{"search":{"q":"\u0633\u0644\u0627\u0645","code":200},"data":{"title":"\u0633\u0644\u0627\u0645","pronunciation":"","text":"(\u0633\u064e) [ \u0639 . ] (\u0645\u0635 \u0644 .) 1 - \u062f\u0631\u0648\u062f \u06af\u0641\u062a\u0646 . 2 - \u0628\u06cc \u06af\u0632\u0646\u062f \u0634\u062f\u0646 . 3 - \u06af\u0631\u062f\u0646 \u0646\u0647\u0627\u062f\u0646 . \u061b ~ \u0639\u0644\u06cc\u06a9 \u062f\u0631\u0648\u062f \u0628\u0631 \u062a\u0648 \u0628\u0627\u062f. \u061b ~ \u0639\u0644\u06cc\u06a9\u0645 \u062f\u0631\u0648\u062f \u0628\u0631 \u0634\u0645\u0627. ","source":"\u0641\u0631\u0647\u0646\u06af \u0641\u0627\u0631\u0633\u06cc \u0645\u0639\u06cc\u0646 | \u0648\u0627\u0698\u0647 \u06cc\u0627\u0628","permalink":"?q=%D8%B3%D9%84%D8%A7%D9%85"},"error":{"message":"","reason":""},"ads":{"text":"","url":""}}

It doesn't seem to have problem ,I can decode it in other tools but PHP, I can't decode it in PHP

$json = file_get_contents('http://api.vajehyab.com/v2/public/?q='.urlencode('سلام').'&developer=stackoverflow');
var_dump(json_decode($json)); // null

I hope somebody could help me with this, thank you...

I did as @Stanimir said, and it's a `- Syntax error, malformed JSON`, — Emad Ghasemi, Jul 01 '15 at 11:06
Works for me on PHP 5.5.21, maybe a question of PHP version then. — ChristianM, Jul 01 '15 at 11:10

score 2 · Answer 1 · answered Jul 01 '15 at 10:48

Taken from php docs http://php.net/manual/bg/function.json-last-error.php Run this code and let us know if you receive any errors.

json_decode($string);

switch(json_last_error())
{
    case JSON_ERROR_DEPTH:
        echo ' - Maximum stack depth exceeded';
    break;
    case JSON_ERROR_CTRL_CHAR:
        echo ' - Unexpected control character found';
    break;
    case JSON_ERROR_SYNTAX:
        echo ' - Syntax error, malformed JSON';
    break;
    case JSON_ERROR_NONE:
        echo ' - No errors';
    break;
}

Hi, it's `- Syntax error, malformed JSON`, :| What should I do now ? :-/ — Emad Ghasemi, Jul 01 '15 at 11:01

score 2 · Answer 2 · edited May 23 '17 at 10:30

2

EDIT2: In PHP 5.3+ this seems to work too:

$json = file_get_contents('http://api.vajehyab.com/v2/public/?q='.urlencode('سلام').'&developer=stackoverflow');

$unescaped = json_encode(preg_replace_callback('/\\\\u(\w{4})/', function ($matches) {
    return html_entity_decode('&#x' . $matches[1] . ';', ENT_COMPAT, 'UTF-8');
}, $json));
var_dump($unescaped);

Taken from: https://stackoverflow.com/a/24933162/2433843

EDIT3: wrapped the preg_replace_callback in json_encode

edited May 23 '17 at 10:30

Community

1
1

answered Jul 01 '15 at 11:06

Francesco de Guytenaere

4,443
2
25
37

No, `json_decode` still return null ! `jsonlint` validate both of them but PHP say it's not :| – Emad Ghasemi Jul 01 '15 at 11:31
Sorry @EmadGhasemi, try this: `$unescaped = json_encode(preg_replace_callback('/\\\\u(\w{4})/', function ($matches) { return html_entity_decode('' . $matches[1] . ';', ENT_COMPAT, 'UTF-8'); }, $json));` – Francesco de Guytenaere Jul 01 '15 at 11:35
You're right, it feels kind of hacky, but it seems it is interpreted as if it was never properly encoded in the first place (I know, it does work in JS Linters). Perhaps it's better to solve it server-side, but perhaps it's a flaw in the PHP json implementation. This seems to work for me though, but I understand if it is too dirty to implement. I'm interested to see the clean solution! – Francesco de Guytenaere Jul 01 '15 at 11:44
It seems that they are using php 5.5.9, and somebody in the comments mentioned it worked with no problem in 5.5.21. Maybe something is going wrong with `json_encode` in 5.5.x :-/ – Emad Ghasemi Jul 01 '15 at 11:50
Ok. Did you try above code? Sorry if it doesn't work; here it works on PHP 5.6.4-4ubuntu6 (local development environment) – Francesco de Guytenaere Jul 01 '15 at 11:53

score 0 · Answer 3 · edited May 23 '17 at 11:47

0

Please check with this.

echo htmlentities((string)$results);

Also reffer Problem with json_decode PHP

edited May 23 '17 at 11:47

Community

1
1

answered Jul 01 '15 at 11:50

Jagadeesh

734
6
26

score 0 · Answer 4 · answered Jul 01 '15 at 14:57

I think the problem is from file_get_contents function. json_decode works fine when input is string. Below code run correctly.

$json = '{"search":{"q":"\u0633\u0644\u0627\u0645","code":200},"data":{"title":"\u0633\u0644\u0627\u0645","pronunciation":"","text":"(\u0633\u064e) [ \u0639 . ] (\u0645\u0635 \u0644 .) 1 - \u062f\u0631\u0648\u062f \u06af\u0641\u062a\u0646 . 2 - \u0628\u06cc \u06af\u0632\u0646\u062f \u0634\u062f\u0646 . 3 - \u06af\u0631\u062f\u0646 \u0646\u0647\u0627\u062f\u0646 . \u061b ~ \u0639\u0644\u06cc\u06a9 \u062f\u0631\u0648\u062f \u0628\u0631 \u062a\u0648 \u0628\u0627\u062f. \u061b ~ \u0639\u0644\u06cc\u06a9\u0645 \u062f\u0631\u0648\u062f \u0628\u0631 \u0634\u0645\u0627. ","source":"\u0641\u0631\u0647\u0646\u06af \u0641\u0627\u0631\u0633\u06cc \u0645\u0639\u06cc\u0646 | \u0648\u0627\u0698\u0647 \u06cc\u0627\u0628","permalink":"?q=%D8%B3%D9%84%D8%A7%D9%85"},"error":{"message":"","reason":""},"ads":{"text":"","url":""}}';
var_dump(json_decode($json));

I tried it with CURL function too, but it doesn't work. – Emad Ghasemi Jul 01 '15 at 15:13 — Emad Ghasemi, Jul 01 '15 at 15:13

score 0 · Answer 5 · answered Jul 26 '16 at 09:41

0

Please encode your data to utf8 format before encoding it through json_encode() php function. I hope , it helps. [utf8_encode()]

answered Jul 26 '16 at 09:41

Lakin Mohapatra

1,097
11
22

Can't decode JSON value in PHP

5 Answers5