PHP json_decode returns null value because of syntax error

Question

$foo_json = '{"hoge":"ywwxEu|`tYdBeARGJ?nJ~BHHmDjX|PdEl@Rj@\\XVtKtK\\VbKnFrIbEzEbBnEbBfDfAxBz@LJjBl@jEzAfZzKDPY|DUdCGbAFL?|AD~AJ\\LNPJ~B|@p@Ll@FRDd@RvBnAfA`@dAl@^NhAVd@@p@?h@H`@Pv@l@TR^f@`Rd@xAFdEVnCLjJn@Cn@FlRDh@HRHHjKpG\\\\jC`G|EvJ|BjGf@vAJh@nGpC`FrBbG`D~@XdBRnNtAnTdBf@Jn@Zf@h@pB`DrD|EZl@jBfEf@`Af@r@tCpChFrG~@zAx@dBdB~Cz@xA`AxA|@l@f@VdDrAnAl@jI~CpCv@|@f@~An@jIxD~CdAzAl@rCvAjI|CzJpClJbCvLdDbn@lH`d@pF~Df@dDn@hA`@hBfA~AbAzQpMvHhFhQ~L~G`FtBrArCnBbBnAjB`A`~@~^tBx@n@RfB\\b@Fh@B|BBpV{AfSgArJq@~@?dBF`Cd@nFjAhXdG~@VtMpEd[fKfL|DxWrI~Y`KvLtDp\\zK|DnAzBbAbA|@rAtApAfBhA`Cl@fBh@lBtEpQh@fBrAjFdDpLdArCj@fAd@x@`BvBlDnDrVxWjDzD`GlGhCvBvBvA|DjBlC|@hTvEfBNvABvISxUo@|JSlA@tBLt@LdAXv@ZnB~@~@t@xAnAf@f@xEdEzAbBzAnBlWp_@bAnBlBpE|@tChBdFhBjGzCnJrCdHp@jAxAxBlGvH~@|Ab@bAp@xBnCdMd@pBl@fBt@bB\\j@v@lAxUlUtCjCdBz@nJdIdJbIrH`H~F|EjDxCzAnBNb@HdAJ|Fb@jDTpHq@@"}';
$foo = json_decode($foo_json);

I wrote this code, but json_decode returns null, and json_last_error() returns "JSON_ERROR_SYNTAX". So I know there are some syntax errors, but I tried the string at JSONLint, they say Valid JSON.

Probably the reason why because the string is complicated and includes symbols. I do not understand what is wrong.

PHP5.6.33

CentOS release 6.6

The answer is here. The problem was backslashes, and I do not like to remove it. json_decode returns JSON_ERROR_SYNTAX but online formatter says the JSON is OK Above article says just remove it, in this case, I need to replace it.

$foo_json = str_replace('\\', '\\\\', $foo_json);

I added this code, then it works fine.

"So I know there are some syntax errors, but I tried the string at JSONLint, they say Valid JSON" — When I try it, it throws up errors. — Quentin, Oct 01 '18 at 07:40
@IslamElshobokshy — Really. Make sure you try to validate your JSON and not the PHP source code for the string literal. — Quentin, Oct 01 '18 at 07:43
Possible duplicate of [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) — Alexandre Elshobokshy, Oct 01 '18 at 07:43

score 1 · Answer 1 · answered Oct 01 '18 at 07:45

1

Look at the rules for strings in JSON:

A \ character has a very limited set of characters it can be followed with.

Your input violates this several times:

e.g. \\X and \\V

Remember that \ is a special character in PHP string literals so if you want an escaped \ in JSON, you need to:

Escape the \ with another \ for JSON
Escape each of them with another \ for PHP

… giving you 4 \ characters.

answered Oct 01 '18 at 07:45

Quentin

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$foo_json = str_replace('\\', '\\\\', $foo_json); Add this code, then works fine! Thank you! – Yuji Hamada Oct 02 '18 at 05:51

score 0 · Answer 2 · answered Oct 01 '18 at 08:02

0

Try this below code

$foo_json =preg_replace('/\\\\/', '', $foo_json);
$foo = json_decode($foo_json);
  print_r( $foo);exit;

It's working fine.

answered Oct 01 '18 at 08:02

VRB

186
14

While this creates JSON that will parse, it discards data from it. The slashes need to be escaped not discarded. – Quentin Oct 01 '18 at 08:08
The problem is backslashes, but I don't like to remove it because I need the backslash data. – Yuji Hamada Oct 01 '18 at 08:19

score -2 · Answer 3 · answered Oct 01 '18 at 07:38

-2

json_(de|en)code needs to have UTF8 data. Check your input if it is UTF8 and if NOT, convert your input to UTF-8 and then use json_(de|en)code to get it ;)

answered Oct 01 '18 at 07:38

Paladin

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Thanks for your answer. This is UTF-8 file, and other json_decode methods succeed with other strings. – Yuji Hamada Oct 01 '18 at 07:44

score -2 · Answer 4 · answered Oct 01 '18 at 07:45

-2

This should result in a valid response. (Handles line breaks and special escapes)

$json = str_replace("\n","\\n",$foo_json );
$json = str_replace("\r","",$json);
$json = preg_replace('/([{,]+)(\s*)([^"]+?)\s*:/','$1"$3":',$json);
$json = preg_replace('/(,)\s*}$/','}',$json);
var_dump($json);

answered Oct 01 '18 at 07:45

Bert Maurau

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5
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Why the downvote? No need for UTF8 encoding or anything.. It's just becase of the escaped values that you need to correctly escape for PHP. The above code parsed your json string correctly.. – Bert Maurau Oct 01 '18 at 07:47
I tested it with the code from the question. The output is still invalid JSON and does not parse. – Quentin Oct 01 '18 at 07:50

PHP json_decode returns null value because of syntax error

4 Answers4