PHP json_decode() returns NULL with seemingly valid JSON?

I have this JSON object stored on a plain text file:

{
    "MySQL": {
        "Server": "(server)",
        "Username": "(user)",
        "Password": "(pwd)",
        "DatabaseName": "(dbname)"
    },
    "Ftp": {
        "Server": "(server)",
        "Username": "(user)",
        "Password": "(pwd)",
        "RootFolder": "(rf)"
    },
    "BasePath": "../../bin/",
    "NotesAppPath": "notas",
    "SearchAppPath": "buscar",
    "BaseUrl": "http:\/\/montemaiztusitio.com.ar",
    "InitialExtensions": [
        "nem.mysqlhandler",
        "nem.string",
        "nem.colour",
        "nem.filesystem",
        "nem.rss",
        "nem.date",
        "nem.template",
        "nem.media",
        "nem.measuring",
        "nem.weather",
        "nem.currency"
    ],
    "MediaPath": "media",
    "MediaGalleriesTable": "journal_media_galleries",
    "MediaTable": "journal_media",
    "Journal": {
        "AllowedAdFileFormats": [
            "flv:1",
            "jpg:2",
            "gif:3",
            "png:4",
            "swf:5"
        ],
        "AdColumnId": "3",
        "RSSLinkFormat": "%DOMAIN%\/notas\/%YEAR%-%MONTH%-%DAY%\/%TITLE%/",
        "FrontendLayout": "Flat",
        "AdPath": "ad",
        "SiteTitle": "Monte Maíz: Tu Sitio",
        "GlobalSiteDescription": "Periódico local de Monte Maíz.",
        "MoreInfoAt": "Más información aquí, en el Periódico local de Monte Maíz.",
        "TemplatePath": "templates",
        "WeatherSource": "accuweather:SAM|AR|AR005|MONTE MAIZ",
        "WeatherMeasureType": "1",
        "CurrencySource": "cotizacion-monedas:Dolar|Euro|Real",
        "TimesSingular": "vez",
        "TimesPlural": "veces"
    }
}

When I try to decode it with json_decode(), it returns NULL. Why? The file is readable (I tried echoing file_get_contents() and it worked ok).

I've tested JSON against http://jsonlint.com/ and it's perfectly valid.

What's wrong here?

Upvotes: 176

Views: 349686

Answers (29)

namal
namal

Reputation: 1274

Before applying PHP related solutions, validate your JSON format. Maybe that is the problem. Try this online JSON format validator.

Upvotes: 0

Dharman
Dharman

Reputation: 33237

This error means that your JSON string is not valid JSON!

Enable throwing exceptions when an error happens and PHP will throw an exception with the reason for why it failed.

Use this:

$json = json_decode($string, null, 512, JSON_THROW_ON_ERROR);

Upvotes: 8

Pekka
Pekka

Reputation: 449435

It could be the encoding of the special characters. You could ask json_last_error() to get definite information.

Upvotes: 97

Muhammad Ichsan
Muhammad Ichsan

Reputation: 425

This happen because you use (') insted {") in your value or key.

Here is wrong format.

{'name':'ichsan'}

Thats will be return NULL if you decode them.

You should pass the json request like this.

{"name":"ichsan"}

Upvotes: 2

Dario Zadro
Dario Zadro

Reputation: 1284

It's probably BOM, as others have mentioned. You can try this:

    // BOM (Byte Order Mark) issue, needs removing to decode
    $bom = pack('H*','EFBBBF');
    $response = preg_replace("/^$bom/", '', $response);
    unset($tmp_bom);
    $response = json_decode($response);

This is a known bug with some SDKs, such as Authorize.NET

Upvotes: 4

devanshi singh
devanshi singh

Reputation: 1

I had the same issue and none of the answers helped me. One of the variables in my JSON object had the value Andaman & Nicobar. I removed this & and my code worked perfectly.

Upvotes: -2

Nim
Nim

Reputation: 129

For me the php function stripslashes() works when receiving json from javascript. When receiving json from python, the second optional parameter to the json_decode call does the trick since the array is associative. Workes for me like a charm.

$json = stripslashes($json); //add this line if json from javascript
$edit = json_decode($json, true); //adding parameter true if json from python

Upvotes: 11

Kolawole Emmanuel Izzy
Kolawole Emmanuel Izzy

Reputation: 1052

Here is how I solved mine https://stackoverflow.com/questions/17219916/64923728 .. The JSON file has to be in UTF-8 Encoding, mine was in UTF-8 with BOM which was adding a weird &65279; to the json string output causing json_decode() to return null

Upvotes: 3

michael aberra
michael aberra

Reputation: 47

In my case , i was facing the same issue , but it was caused by slashes inside the json string so using

json_decode(stripslashes($YourJsonString))

OR

json_decode( preg_replace('/[\x00-\x1F\x80-\xFF]/', '', $YourJsonString), true );

If the above doesnt work, first replace the quotes from html quote , this might be happening if you are sending data from javascript to php

    $YourJsonString = stripslashes($YourJsonString);
    $YourJsonString = str_replace('"', '"', $YourJsonString);
    $YourJsonString = str_replace('["', '[', $YourJsonString);
    $YourJsonString = str_replace('"]', ']', $YourJsonString);
    $YourJsonString = str_replace('"{', '{', $YourJsonString);
    $YourJsonString = str_replace('}"', '}', $YourJsonString);
    $YourJsonObject = json_decode($YourJsonString);

Will solve it,

Upvotes: 2

hasan akbari
hasan akbari

Reputation: 5

I had exactly the same problem But it was fixed with this code

$zip = file_get_contents($file);
$zip = json_decode(stripslashes($zip), true);

Upvotes: -2

Ashish Sondagar
Ashish Sondagar

Reputation: 1095

  • I also face the same issue...
  • I fix the following steps... 1) I print that variable in browser 2) Validate that variable data by freeformatter 3) copy/refer that data in further processing
  • after that, I didn't get any issue.

Upvotes: 2

user419685
user419685

Reputation: 11

I recommend creating a .json file (ex: config.json). Then paste all of your json object and format it. And thus you will be able to remove all of that things that is breaking your json-object, and get clean copy-paste json-object.

Upvotes: 1

NeoTechni
NeoTechni

Reputation: 179

It took me like an hour to figure it out, but trailing commas (which work in JavaScript) fail in PHP.
This is what fixed it for me:

str_replace([PHP_EOL, ",}"], ["", "}"], $JSON);

Upvotes: 2

Suman Deol
Suman Deol

Reputation: 384

So, html_entity_decode() worked for me. Please try this.

$input = file_get_contents("php://input");
$input = html_entity_decode($input);
$event_json = json_decode($input,true);

Upvotes: 2

Grant
Grant

Reputation: 6329

The most important thing to remember, when you get a NULL result from JSON data that is valid is to use the following command:

json_last_error_msg();

Ie.

var_dump(json_last_error_msg());
string(53) "Control character error, possibly incorrectly encoded"

You then fix that with:

$new_json = preg_replace('/[[:cntrl:]]/', '', $json);

Upvotes: 3

shasi kanth
shasi kanth

Reputation: 7094

For me, I had to turn off the error_reporting, to get json_decode() working correctly. It sounds weird, but true in my case. Because there is some notice printed between the JSON string that I am trying to decode.

Upvotes: -2

Tho
Tho

Reputation: 25070

For my case, it's because of the single quote in JSON string.

JSON format only accepts double-quotes for keys and string values.

Example:

$jsonString = '{\'hello\': \'PHP\'}'; // valid value should be '{"hello": "PHP"}'
$json = json_decode($jsonString);
print $json; // null

I got this confused because of Javascript syntax. In Javascript, of course, we can do like this:

let json = {
    hello: 'PHP' // no quote for key, single quote for string value
}

// OR:
json = {
    'hello': 'PHP' // single quote for key and value
}

but later when convert those objects to JSON string:

JSON.stringify(json); // "{"hello":"PHP"}"

Upvotes: 0

Hassan Saeed
Hassan Saeed

Reputation: 7090

you should ensure these points

1. your json string dont have any unknowns characters

2. json string can view from online json viewer (you can search on google as online viewer or parser for json) it should view without any error

3. your string dont have html entities it should be plain text/string

for explanation of point 3

$html_product_sizes_json=htmlentities($html);
    $ProductSizesArr = json_decode($html_product_sizes_json,true);

to (remove htmlentities() function )

$html_product_sizes_json=$html;
    $ProductSizesArr = json_decode($html_product_sizes_json,true);

Upvotes: 0

Enrico Tempesti
Enrico Tempesti

Reputation: 125

this help you to understand what is the type of error

<?php
// A valid json string
$json[] = '{"Organization": "PHP Documentation Team"}';

// An invalid json string which will cause an syntax 
// error, in this case we used ' instead of " for quotation
$json[] = "{'Organization': 'PHP Documentation Team'}";


foreach ($json as $string) {
    echo 'Decoding: ' . $string;
    json_decode($string);

    switch (json_last_error()) {
        case JSON_ERROR_NONE:
            echo ' - No errors';
        break;
        case JSON_ERROR_DEPTH:
            echo ' - Maximum stack depth exceeded';
        break;
        case JSON_ERROR_STATE_MISMATCH:
            echo ' - Underflow or the modes mismatch';
        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_UTF8:
            echo ' - Malformed UTF-8 characters, possibly incorrectly encoded';
        break;
        default:
            echo ' - Unknown error';
        break;
    }

    echo PHP_EOL;
}
?>

Upvotes: 6

Jeffrey Roosendaal
Jeffrey Roosendaal

Reputation: 7157

I've solved this issue by printing the JSON, and then checking the page source (CTRL/CMD + U):

print_r(file_get_contents($url));

Turned out there was a trailing <pre> tag.

Upvotes: 0

Albert Abdonor
Albert Abdonor

Reputation: 961

Maybe some hidden characters are messing with your json, try this:

$json = utf8_encode($yourString);
$data = json_decode($json);

Upvotes: 27

user2254008
user2254008

Reputation: 305

If you check the the request in chrome you will see that the JSON is text, so there has been blank code added to the JSON.

You can clear it by using

$k=preg_replace('/\s+/', '',$k);

Then you can use:

json_decode($k)

print_r will then show the array.

Upvotes: 27

Gabriel Castillo Prada
Gabriel Castillo Prada

Reputation: 4673

You could try with it.

json_decode(stripslashes($_POST['data']))

Upvotes: 64

Samuel Kwame Antwi
Samuel Kwame Antwi

Reputation: 597

Just save some one time. I spent 3 hours to find out that it was just html encoding problem. Try this

if(get_magic_quotes_gpc()){
   $param = stripslashes($row['your column name']);
}else{
  $param = $row['your column name'];
}

$param = json_decode(html_entity_decode($param),true);
$json_errors = array(
JSON_ERROR_NONE => 'No error has occurred',
JSON_ERROR_DEPTH => 'The maximum stack depth has been exceeded',
JSON_ERROR_CTRL_CHAR => 'Control character error, possibly incorrectly encoded',
JSON_ERROR_SYNTAX => 'Syntax error',
);
echo 'Last error : ', $json_errors[json_last_error()], PHP_EOL, PHP_EOL;
print_r($param);

Upvotes: 1

TomoMiha
TomoMiha

Reputation: 1279

If you are getting json from database, put

mysqli_set_charset($con, "utf8");

after defining connection link $con

Upvotes: 2

Dunith Dhanushka
Dunith Dhanushka

Reputation: 4334

This worked for me

json_decode( preg_replace('/[\x00-\x1F\x80-\xFF]/', '', $json_string), true );

Upvotes: 139

Charles Yapp
Charles Yapp

Reputation: 566

I had the same problem and I solved it simply by replacing the quote character before decode.

$json = str_replace('&quot;', '"', $json);
$object = json_decode($json);

My JSON value was generated by JSON.stringify function.

Upvotes: 17

Phil LaNasa
Phil LaNasa

Reputation: 3035

Just thought I'd add this, as I ran into this issue today. If there is any string padding surrounding your JSON string, json_decode will return NULL.

If you're pulling the JSON from a source other than a PHP variable, it would be wise to "trim" it first:

$jsonData = trim($jsonData);

Upvotes: 6

user2648057
user2648057

Reputation: 7

<?php 
$json_url = "http://api.testmagazine.com/test.php?type=menu";
$json = file_get_contents($json_url);
$json=str_replace('},

]',"}

]",$json);
$data = json_decode($json);

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

Upvotes: -6

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