PHP: json decode limits

Question

Take this code:

$json = file_get_contents($this->url, true); 
$decode = json_decode($json, true); 

foreach ($decode as $key => $value) {
 ...
}

Pretty simple, uh?

Passing a $json with up to 500 array elements.... works right!

Above that limit... the error is:

Warning: Invalid argument supplied for foreach() in /c/website/retriever/WsGlassRetriever.php on line 19

Is there some memory limit for that function's argument?

I didn't found nothing about it in the docs. My version is PHP 5.2.17-rnx1.1 with Suhosin-Patch 0.9.7 (cli)

i dont think its limit. Its not limit. try var_dump($decode) to see type — safarov
– safarov, Commented Mar 28, 2012 at 13:24
it may be not because of the count of the elements, but because of particular element that can't be decoded. In php 5.3 there is json_last_error() function. Also mind the depth limit (512 by default) — Maxim Krizhanovsky
– Maxim Krizhanovsky, Commented Mar 28, 2012 at 13:24
You can use json_last_error() [edit: requires PHP >= 5.3!] to determine what exactly went wrong: php.net/manual/en/function.json-last-error.php — dbrumann
– dbrumann, Commented Mar 28, 2012 at 13:25
@Darhazer 512 now, but used to be 128 and before that used to be 20. If your pre-5.2.3 this can be a real problem, but the OP is 5.2.17 so his is 128. — DaveRandom
– DaveRandom, Commented Mar 28, 2012 at 13:29

Niet the Dark Absol · Accepted Answer · 2012-03-28 13:27:57Z

15

json_decode returns NULL if there is an error in the JSON syntax. I've just successfully tested on an array of 1000 elements and it ran just fine.

Double-check that your JSON is correctly formatted. Even something as small as having single quotes instead of double, or forgetting to put a property name in quotes, or using a character outside the 32-127 range without correctly encoding it in UTF-8 can cause these problems.

answered Mar 28, 2012 at 13:27

Niet the Dark Absol

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2 Comments

Fabio B. Over a year ago

Ok. it was a malformed utf-8 error. I looked at my json and there I found a 'ì' char which was bad utf8... What about validating my utf-8 json on the fly with php?

TimSparrow Over a year ago

I use GuzzleHttp's json_decode wrapper, it throws an InvalidArgumentException when JSON data is bad

Baba · Accepted Answer · 2012-03-28 13:53:33Z

Am sure your JSON code above 500 has a formatting issue , have used JSON with over 20,000 values here is a simple script of 2000 array

$string = "Sample String Data ¶";
$string = preg_replace( '/[^[:print:]]/', '',$string); // remove all values that can affect JSON 
$array = array();
for($i = 0 ; $i <  2000; $i++)
{
    if(mt_rand(0, 1))
    {
        $array[] = $string ;
    }
    else
    {
        $array[] = array($string,1,$string) ;
    }
}   

$json =  json_encode($array);
$decodeArray =  json_decode($json);

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 "<br />" ;


foreach ($decodeArray as $key => $value) {
    print_r($value) ;
    flush();
}

Edit 2

I was so interested to know if there is any limitation .. just tested it with 250,000 (Two hundred and fifty thousand values and it works fine )

Thanks Oleku

wellsantos · Accepted Answer · 2017-11-07 16:53:47Z

0

In my case the JSON was correct. My problem was the "JSON_BIGINT_AS_STRING" parameter that was causing the error "Maximum stack depth exceeded".

$jsonResult = json_decode($expr,true,JSON_BIGINT_AS_STRING);

I removed the "JSON_BIGINT_AS_STRING" argument and the error is gone:

$jsonResult = json_decode($expr,true);

answered Nov 7, 2017 at 16:53

wellsantos

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1 Comment

Niklaus Over a year ago

Your problem might have been the fact that you defined max depth with JSON_BIGINT_AS_STRING, which equals 2. You should have done $jsonResult = json_decode($expr, true, 512, JSON_BIGINT_AS_STRING);

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