PHP json_decode gives empty array

Question

Hi I'm trying to decode a string with json format into a associative array. The string is this one: (My string is from a database and its generated there)

{ 
  "Parameter1":"<style>
                  #label-9 {
                  display: block;
                  text-align: left;
                  color: #fff;
                  }
                </style>", 
  "HistoryPosition":"1"
}

And when I do json_decode() it gives me an empty array. Do you know why this happens? I believe it's something from the "Parameter1" but can't find what it is.

Thank you :)

Try json_last_error_msg and see if that shows any problems. — Niet the Dark Absol
– Niet the Dark Absol, Commented May 27, 2016 at 14:49
I do believe it is because of the line breaks, JSON does not allow real line-breaks — Akshay
– Akshay, Commented May 27, 2016 at 14:49
Akshay is right. see json.org/string.gif , form-feed,new-line,carriage-return are all control characters and therefore not allowed in the string literal. — VolkerK
– VolkerK, Commented May 27, 2016 at 14:56
@Akshay is there a way to convert this to an associative array without loosing the line breaks of the string? — patricia
– patricia, Commented May 27, 2016 at 14:58
@patricia sorry, that was a typo. You need to replace the linebreaks in value of Parameter1 with \n, so for example the first parameter would look like this: <style>\n #label-9. That way you'll preserve the line breaks when you e.g. pull it from a database and output it into a webpage. You should expand your question though, as it's unclear if you have access to the string as-is, or if you have access to individual values as well. — MicE
– MicE, Commented May 27, 2016 at 15:20

Machavity · Accepted Answer · 2016-05-27 14:58:44Z

1

JSONLint indicates the JSON is invalid.

What you might want to do is the following

$json = '{ 
  "Parameter1":"<style>
                  #label-9 {
                  display: block;
                  text-align: left;
                  color: #fff;
                  }
                </style>", 
  "HistoryPosition":"1"
}';
// remove the newlines
$clean = str_replace(["\r", "\n"], ['', ''], $json);
var_dump(json_decode($clean));

Demo https://3v4l.org/X6KgM

answered May 27, 2016 at 14:58

Machavity♦

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3 Comments

patricia Over a year ago

Is there a way to get back the line breaks after decoding?

Machavity Over a year ago

The problem is that you have line breaks outside the quoted area too. So you can't just replace the line breaks with JS line breaks, or else you'll wind up with invalid JSON again. You could maybe try str_replace(';', ";\n", $yourvar->Parameter1); to put some of them back

patricia Over a year ago

I could do it since in my database when I generate this string I don't put line breaks outside the quoted area. (I just put it here so the information is readable.

MicE · Accepted Answer · 2016-05-27 14:57:54Z

Akshay is indeed correct, it is caused by linebreaks.

<pre><?php

$input = <<<EOD
{
  "Parameter1":"<style>
                  #label-9 {
                  display: block;
                  text-align: left;
                  color: #fff;
                  }
                </style>",
  "HistoryPosition":"1"
}
EOD;

// json_decode($input, true);
// echo json_last_error_msg(); // Syntax error

$input = str_replace("\r", null, $input);
$input = str_replace("\n", null, $input);

var_dump(json_decode($input, true));

Prints:

array(2) {
  ["Parameter1"]=> string(176) "<style>                  #label-9 {                  display: block;                  text-align: left;                  color: #fff;                  }                </style>"
  ["HistoryPosition"]=> string(1) "1"
}

MonkeyZeus · Accepted Answer · 2016-05-27 15:19:58Z

1

Instead of handwriting your own JSON string, you should absolutely be using PHP's built-in functions to make your like at least 483% easier:

// Use a native PHP array to store your data; it will preserve the new lines
$input = [
    "Parameter1" => "<style>
                  #label-9 {
                  display: block;
                  text-align: left;
                  color: #fff;
                  }
                </style>",
    "HistoryPosition" => "1"
];

// This function will preserve everything in your strings
$encoded_value = json_encode($input);

// See your properly formatted JSON string
echo $encoded_value.'<br><br>';

// Decode the string back into an associative PHP array
echo '<pre>';
print_r(json_decode($encoded_value, true));
echo '</pre>';

Update per new info about DB retrieval

json_last_error_msg(); produces this error:

Control character error, possibly incorrectly encoded

If you do not care about preserving newlines then this will work:

<?php
$db_data = '{ 
  "Parameter1":"<style>
                  #label-9 {
                  display: block;
                  text-align: left;
                  color: #fff;
                  }
                </style>", 
  "HistoryPosition":"1"
}';

$db_data = str_replace("\r", "", $db_data);
$db_data = str_replace("\n", "", $db_data);

echo '<pre>';
print_r(json_decode($db_data, true));
echo '</pre>';

edited May 27, 2016 at 15:19

answered May 27, 2016 at 15:09

MonkeyZeus

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1 Comment

patricia Over a year ago

I can't do that because the data is generated and saved in the database.

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