Php json string object to php array

Question

I've looked at how to decode a string here however it requires the key's be wrapped in quotes and my data doesnt, example:

The below data, is saved in a .txt file, I'm using file_get_contents() to read the files, I do not have control over the below data.

THIS IS MY DATA

"{
    ip : "192.168.1.110",
    startFrame : "1",
    endFrame : "11",
    startedCurrentFrameAt: "1397529891",
    status: "rendering",
    currentFrame: "0"
}"

In php, I want to be able to read this data, and access each key, this is what I've tried:

$arr = json_decode($data, true)['status'];

$arr just returns null, because the key's aren't quoted, is there a work around for this?

I've found many answers to this question, but all have the key's quoted.

Are you generating the JSON or are you getting it from somewhere ? — Shankar Narayana Damodaran
– Shankar Narayana Damodaran, Commented Apr 15, 2014 at 4:21
From 2.2. Objects An object structure is represented as a pair of curly brackets surrounding zero or more name/value pairs (or members). A name is a string. and from 2.5. Strings A string begins and ends with quotation marks. So those would technically be objects, which json does'nt support. the keys need to be strings my friend — Michael King
– Michael King, Commented Apr 15, 2014 at 4:24
@ShankarDamodaran I have edited my question, I'm basically reading a txt file which the data is automatically populated. — Shannon Hochkins
– Shannon Hochkins, Commented Apr 15, 2014 at 4:26
@AwladLiton I'm not generating the object, it's automatically pulled in through a .txt which I do not create/manage. — Shannon Hochkins
– Shannon Hochkins, Commented Apr 15, 2014 at 4:27

meda · Accepted Answer · 2014-04-15 05:05:41Z

2

Try this

<?php
 function fix_json($s) {
  $s = preg_replace('/(\w+):/i', '"\1":', $s);
  return $s;
}


$data = '{
    ip: "192.168.1.110",
    startFrame: "1",
    endFrame: "11",
    startedCurrentFrameAt: "1397529891",
    status: "rendering",
    currentFrame: "0"
}';

$valid_json = fix_json($data);
$arr = json_decode($valid_json , true);
$status = $arr['status'];

echo $status;

DEMO

edited Apr 15, 2014 at 5:05

answered Apr 15, 2014 at 4:30

meda

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3 Comments

Shannon Hochkins Over a year ago

Please3 read my whole question, I do not have control over the quotes around the keys, unless I use a regex to add them.

Shannon Hochkins Over a year ago

My object is wrapped in "" not '', why does this matter? It doesn't seem to work unless it's single quotes not double

Shannon Hochkins Over a year ago

I eventually got this working, it wasn't work because I had spaces in my datakey : "s", I used another expression to remove the any spaces before a semi colon and it worked fine

Community · Accepted Answer · 2020-06-20 09:12:55Z

1

Use `preg_replace_callback()` for this.

Well what is happening in the code ?

First, the regular expression is trying to find the entries between a space and a : , and then concatenates the quotes around them. Finally, an str_replace() acts a wrapper to fix the JSON braces.

<?php

$json='"{
    ip : "192.168.1.110",
    startFrame : "1",
    endFrame : "11",
    startedCurrentFrameAt: "1397529891",
    status: "rendering",
    currentFrame: "0"
}"';

function cb_quote($v)
{
    return '"'.trim($v[1]).'":';
}
$newJSON=str_replace(array('"{','}"'),array('[{','}]'),preg_replace_callback("~ (.*?):~","cb_quote", $json));
echo $arr = json_decode($newJSON, true)[0]['status'];

OUTPUT :

rendering

Working Demo

Through a file.. (Edit)

<?php
$json = trim(file_get_contents('new.txt'));
//Modifications..
$json = str_replace(array('{','}',':',','),array('[{" ',' }]','":',',"'),$json);
function cb_quote($v)
{
    return '"'.trim($v[1]).'"';
}
$newJSON=preg_replace_callback("~\"(.*?)\"~","cb_quote", $json);
echo $arr = json_decode($newJSON, true)[0]['status']; //"prints" rendering

edited Jun 20, 2020 at 9:12

CommunityBot

11 silver badge

answered Apr 15, 2014 at 5:10

Shankar Narayana Damodaran

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8 Comments

Shannon Hochkins Over a year ago

I am still getting null with this, I can't define the $json variable like you have, to try and replicate, would I just write this $json = "'" . $data ."'"; ??

Shankar Narayana Damodaran Over a year ago

If you are getting that from a text file , simply do ... $json = trim(file_get_contents('yourjsonfile.txt'));

Shannon Hochkins Over a year ago

newJSON is defined, but if I just return $arr like so: return json_decode($newJSON, true); it is null...

Shankar Narayana Damodaran Over a year ago

Why in the world are you using return? Are you inside a function or what ?

Shannon Hochkins Over a year ago

I beleive it's formatting my string wrong, I've just seen in newJSON it's being formatted like so:

"{ip"":""192.168.1.110",startFrame":""1",endFrame":""11",startedCurrentFrameAt":""1397529891",status":""rendering",currentFrame":""0"}"

|

Ryan · Accepted Answer · 2014-04-15 04:46:49Z

0

I would do this by using the PHP explode function. Explode is like javascript's .split function, in that it will take a string and EXPLODE it into an array of usable data.

So, something like this should work, but you'll need to remove the "{ and }" on the ends first:

// lets assume that your data is $mydata, for brevity's sake.
$mydata_arr = new Array();
$item_arr = new Array();
$newdata_arr = new Array();

$mydata_arr = explode(',',$mydata); // explode where there is a comma

foreach($item in $mydata_arr){
    $item_arr = explode(':',$item); // explode where there is a colon
    // this gives us keys and values in a the $item_arr at [0] and [1]
    $key = $item_arr[0];
    $value = $item_arr[1]; 

    // reassign these in a new, usable array     
    $newdata_arr[$key] = $value;
}

Now, I know this is crude and inelegant, but I wanted to explain in a way that would help you see what needs to be done in something like English. I hope it helps.

answered Apr 15, 2014 at 4:46

Ryan

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2 Comments

Shannon Hochkins Over a year ago

I don't know where, but I believe there's an error somewhere, I can't get this to work :(

Ryan Over a year ago

Have you removed the starting and ending bits? That would mess it up.

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Php json string object to php array

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Use `preg_replace_callback()` for this.

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Use preg_replace_callback() for this.

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Use `preg_replace_callback()` for this.