PHP json_decode return error code 4

Question

I had previously asked the same question. I would like to decode the json from: http://pad.skyozora.com/data/pets.json. Below is the code I used previously:

<?php
$html=file_get_contents("http://pad.skyozora.com/data/pets.json");
var_dump(json_decode($html,true)); //return null
var_dump(json_last_error()); // return 4
?>

From the last answer I know there is UTF8 DOM in the json return. I tried the answer from a similar question: json_decode returns NULL after webservice call, but all of the answers not work for me.

And after do more research I found a way that works:

<?php
$html=file_get_contents("http://pad.skyozora.com/data/pets.json");
$html=preg_replace('/[\x00-\x1F\x80-\xFF]/', '', $html);
var_dump(json_decode($html, true));
var_dump(json_last_error());
?>

This successfully decode the json into array. However all Chinese and Japanese character string were removed too. Any ideas?

Edited:

I used http://jsonlint.com/ to decode the json from http://pad.skyozora.com/data/pets.json. It stops at here:

[
            161,
            "進化的紅面具",
            0,
            -1,
            0,
        -1,
        1,
        1,
        10,
        50,
        1,
        0,
        0,
        0,
        0,
        [

        ],
        [
            0,
            0,
            0,
            0,
            0,
            0,
            0
        ],
        "http:\/\/i.imgur.com\/Y1jZlGW.png",
        [
            "ウルカヌ火山",
            "メジェド

and give me the error:

Parse error on line 5001:
...山",                "メジェド
----------------------^
Expecting 'STRING', 'NUMBER', 'NULL', 'TRUE', 'FALSE', '{', '['

But I couldn't see any problem with this part of json.

Does this answer your question? [json\_decode returns JSON\_ERROR\_SYNTAX but online formatter says the JSON is OK](https://stackoverflow.com/questions/17219916/json-decode-returns-json-error-syntax-but-online-formatter-says-the-json-is-ok) — Ooker, Jan 13 '22 at 09:04
For me NaN was the problem, using df.fillna('', inplace=True) fix the problem in case if you used pandas — Khaled Alam, May 14 '23 at 23:20

Adrian Cid Almaguer · Answer 1 · 2015-03-29T06:13:47.253

10

The return error code 4 is due to a JSON_ERROR_SYNTAX you must try to fix your json.

If you go to the url in a browser you can see this message:

bad character in string literal at line 1 column 294388

edited Mar 29 '15 at 06:13

answered Mar 29 '15 at 06:08

Adrian Cid Almaguer

7,815
13
41
63

1

Thanks for your answer, how to resolve the syntax error? – Patrick Ng Mar 29 '15 at 06:18
@NgYatPing Well this is a hard road, first search where it fail and post the code. – Adrian Cid Almaguer Mar 29 '15 at 06:20
I have edited the question with validator result from http://jsonlint.com/ . But I don't see any problem with it.... – Patrick Ng Mar 29 '15 at 06:41
@NgYatPing Change the value where the validator stop and see waht happens – Adrian Cid Almaguer Mar 29 '15 at 07:02
@NgYatPing you got some advance? – Adrian Cid Almaguer Mar 31 '15 at 19:40
Hi Adrian, finally I still cannot get to decode the json through json_decode(). I used http://jsonformat.com/#jsondataurllabel to help me decode instead... And I still don't know the reason... – Patrick Ng Apr 01 '15 at 16:42

PHP json_decode return error code 4

1 Answers1

Linked

Related