PHP reading invalid json with json_decode();

Question

I have invalid external json data, without double quotes around names.

Example:

{
  data: [
    {
      idx: 0,
      id: "0",
      url: "http://247wallst.com/",
      a: [
        {
          t: "Title",
          u: "http://247wallst.com/2012/07/30/",
          sp: "About"
        }
      ],
      doc_id: "9386093612452939480"
    },
    {
      idx: 1,
      id: "-1"
    }
  ],
  results_per_page: 10,
  total_number_of_news: 76,
  news_per_month: [20, 0, 8, 1, 1, 2, 0, 2, 1, 0, 0, 1, 1, 0, 5, 1, 1, 1, 0, 2, 5, 16, 7, 1],
  result_start_num: 2,
  result_end_num: 2,
  result_total_articles: 76
}

As you see a lot of names like data,idx,id,url and others are not double quoted, so this makes this json invalid. How can I make this external json valid? I already tried str_replace, replacing '{' to '{"' and ':' to '":' adding double quotes around unquoted names, but this messes up some already double quoted variables.

How can I make this json valid so I can read this data with PHP json_decode? I'm not very familiar with preg_replace..

Valid json will look like:

{
  "data": [
    {
      "idx": 0,
      "id": "0",
      "url": "http://247wallst.com/",
      "a": [
        {
          "t": "Title",
          "u": "http://247wallst.com/2012/07/30/",
          "sp": "About"
        }
      ],
      "doc_id": "9386093612452939480"
    },
    {
      "idx": 1,
      "id": "-1"
    }
  ],
  "results_per_page": 10,
  "total_number_of_news": 76,
  "news_per_month": [20, 0, 8, 1, 1, 2, 0, 2, 1, 0, 0, 1, 1, 0, 5, 1, 1, 1, 0, 2, 5, 16, 7, 1],
  "result_start_num": 2,
  "result_end_num": 2,
  "result_total_articles": 76
}

Please suggest me some php preg_replace function.

Data source: http://www.google.com/finance/company_news?q=aapl&output=json&start=1&num=1

Answer 1

With preg_replace you can do:

json_decode(preg_replace('#(?<pre>\{|\[|,)\s*(?<key>(?:\w|_)+)\s*:#im', '$1"$2":', $in));

Since the above example won't work with real data (the battle plans seldom survive first contact with the enemy) heres my second take:

$infile = 'http://www.google.com/finance/company_news?q=aapl&output=json&start=1&num=1';

// first, get rid of the \x26 and other encoded bytes.
$in = preg_replace_callback('/\\\x([0-9A-F]{2})/i',
    function($match){
        return chr(intval($match[1], 16));
    }, file_get_contents($infile));

$out = $in;

// find key candidates
preg_match_all('#(?<=\{|\[|,)\s*(?<key>(?:\w|_)+?)\s*:#im', $in, $m, PREG_OFFSET_CAPTURE);

$replaces_so_far = 0;
// check each candidate if its in a quoted string or not
foreach ($m['key'] as $match) {
    $position = $match[1] + ($replaces_so_far * 2); // every time you expand one key, offsets need to be shifted with 2 (for the two " chars)
    $key = $match[0];
    $quotes_before = preg_match_all('/(?<!\\\)"/', substr($out, 0, $position), $m2);
    if ($quotes_before % 2) { // not even number of not-escaped quotes, we are in quotes, ignore candidate
        continue;
    }
    $out = substr_replace($out, '"'.$key.'"', $position, strlen($key));
    ++$replaces_so_far;
}

var_export(json_decode($out, true));

But since google offers this data in RSS feed, i would recommend you to use that one if it works for your usecase, this is just for fun (-:

Answer 2

The JSON feeds from Google always seem to be plagued with problems- formatted incorrectly in some way shape or form. If you switch the feed to RSS you can easily convert it to an array or JSON from the array.

<?php

$contents = file_get_contents('http://www.google.com/finance/company_news?q=aapl&output=rss&start=1&num=1');

// Convert the RSS to an array (probably just use this)
$arr = simplexml_load_string($contents);

// Or if you specifically want JSON
$json = json_encode($arr);

// And back to an array
print_r(json_decode($json));