How to convert a string to JSON object in PHP

Question

I have the following result from an SQL query:

{"Coords":[
    {"Accuracy":"65","Latitude":"53.277720488429026","Longitude":"-9.012038778269686","Timestamp":"Fri Jul 05 2013 11:59:34 GMT+0100 (IST)"},
    {"Accuracy":"65","Latitude":"53.277720488429026","Longitude":"-9.012038778269686","Timestamp":"Fri Jul 05 2013 11:59:34 GMT+0100 (IST)"},
    {"Accuracy":"65","Latitude":"53.27770755361785","Longitude":"-9.011979642121824","Timestamp":"Fri Jul 05 2013 12:02:09 GMT+0100 (IST)"},
    {"Accuracy":"65","Latitude":"53.27769091555766","Longitude":"-9.012051410095722","Timestamp":"Fri Jul 05 2013 12:02:17 GMT+0100 (IST)"},
    {"Accuracy":"65","Latitude":"53.27769091555766","Longitude":"-9.012051410095722","Timestamp":"Fri Jul 05 2013 12:02:17 GMT+0100 (IST)"} 
    ]
}

It is currently a string in PHP. I know it's already in JSON form, is there an easy way to convert this to a JSON object?

I need it to be an object so I can add an extra item/element/object like what "Coords" already is.

@user2363025 this is your string converted to valid JSON: http://pastebin.com/R16NVerw — Miro Markaravanes, Jul 05 '13 at 11:58
@MiroMarkarian Although the JSON is valid, you broke the date format xD — Dan, Jul 05 '13 at 12:09
The JsonLint did that I didn't. I posted just to show what is valid. When converting programatically, it won't happen — Miro Markaravanes, Jul 05 '13 at 12:28
I saw that you edited the question, tell me: the initial data are in JSON format you want to convert to PHP variables (Array/Object - stdClass), add data and convert to JSON again? Note: I edited my answer. — Protomen, Jul 06 '13 at 00:00

score 168 · Accepted Answer · edited Dec 26 '22 at 06:40

What @deceze said is correct, it seems that your JSON is malformed, try this:

{
    "Coords": [{
        "Accuracy": "30",
        "Latitude": "53.2778273",
        "Longitude": "-9.0121648",
        "Timestamp": "Fri Jun 28 2013 11:43:57 GMT+0100 (IST)"
    }, {
        "Accuracy": "30",
        "Latitude": "53.2778273",
        "Longitude": "-9.0121648",
        "Timestamp": "Fri Jun 28 2013 11:43:57 GMT+0100 (IST)"
    }, {
        "Accuracy": "30",
        "Latitude": "53.2778273",
        "Longitude": "-9.0121648",
        "Timestamp": "Fri Jun 28 2013 11:43:57 GMT+0100 (IST)"
    }, {
        "Accuracy": "30",
        "Latitude": "53.2778339",
        "Longitude": "-9.0121466",
        "Timestamp": "Fri Jun 28 2013 11:45:54 GMT+0100 (IST)"
    }, {
        "Accuracy": "30",
        "Latitude": "53.2778159",
        "Longitude": "-9.0121201",
        "Timestamp": "Fri Jun 28 2013 11:45:58 GMT+0100 (IST)"
    }]
}

Use json_decode to convert String into Object (stdClass) or array: http://php.net/manual/en/function.json-decode.php

[edited]

I did not understand what do you mean by "an official JSON object", but suppose you want to add content to json via PHP and then converts it right back to JSON?

assuming you have the following variable:

$data = '{"Coords":[{"Accuracy":"65","Latitude":"53.277720488429026","Longitude":"-9.012038778269686","Timestamp":"Fri Jul 05 2013 11:59:34 GMT+0100 (IST)"},{"Accuracy":"65","Latitude":"53.277720488429026","Longitude":"-9.012038778269686","Timestamp":"Fri Jul 05 2013 11:59:34 GMT+0100 (IST)"},{"Accuracy":"65","Latitude":"53.27770755361785","Longitude":"-9.011979642121824","Timestamp":"Fri Jul 05 2013 12:02:09 GMT+0100 (IST)"},{"Accuracy":"65","Latitude":"53.27769091555766","Longitude":"-9.012051410095722","Timestamp":"Fri Jul 05 2013 12:02:17 GMT+0100 (IST)"},{"Accuracy":"65","Latitude":"53.27769091555766","Longitude":"-9.012051410095722","Timestamp":"Fri Jul 05 2013 12:02:17 GMT+0100 (IST)"}]}';

You should convert it to Object (stdClass):

$manage = json_decode($data);

But working with stdClass is more complicated than PHP-Array, then try this (use second param with true):

$manage = json_decode($data, true);

This way you can use array functions: http://php.net/manual/en/function.array.php

adding an item:

$manage = json_decode($data, true);

echo 'Before: <br>';
print_r($manage);

$manage['Coords'][] = Array(
    'Accuracy' => '90'
    'Latitude' => '53.277720488429026'
    'Longitude' => '-9.012038778269686'
    'Timestamp' => 'Fri Jul 05 2013 11:59:34 GMT+0100 (IST)'
);

echo '<br>After: <br>';
print_r($manage);

remove first item:

$manage = json_decode($data, true);
echo 'Before: <br>';
print_r($manage);
array_shift($manage['Coords']);
echo '<br>After: <br>';
print_r($manage);

any chance you want to save to json to a database or a file:

$data = '{"Coords":[{"Accuracy":"65","Latitude":"53.277720488429026","Longitude":"-9.012038778269686","Timestamp":"Fri Jul 05 2013 11:59:34 GMT+0100 (IST)"},{"Accuracy":"65","Latitude":"53.277720488429026","Longitude":"-9.012038778269686","Timestamp":"Fri Jul 05 2013 11:59:34 GMT+0100 (IST)"},{"Accuracy":"65","Latitude":"53.27770755361785","Longitude":"-9.011979642121824","Timestamp":"Fri Jul 05 2013 12:02:09 GMT+0100 (IST)"},{"Accuracy":"65","Latitude":"53.27769091555766","Longitude":"-9.012051410095722","Timestamp":"Fri Jul 05 2013 12:02:17 GMT+0100 (IST)"},{"Accuracy":"65","Latitude":"53.27769091555766","Longitude":"-9.012051410095722","Timestamp":"Fri Jul 05 2013 12:02:17 GMT+0100 (IST)"}]}';

$manage = json_decode($data, true);

$manage['Coords'][] = Array(
    'Accuracy' => '90'
    'Latitude' => '53.277720488429026'
    'Longitude' => '-9.012038778269686'
    'Timestamp' => 'Fri Jul 05 2013 11:59:34 GMT+0100 (IST)'
);

if (($id = fopen('datafile.txt', 'wb'))) {
    fwrite($id, json_encode($manage));
    fclose($id);
}

Thank you been pulling out my hair as to why the array I am pushing to another array is being shown as a string instead of an object. — SleepNot, Dec 10 '14 at 06:12
If you're not dealing with `stdClass` and it's not working with the simple `json_decode` function, by casting the string as an `(array)` beforehand did the trick for me. — LatentDenis, Jun 07 '17 at 17:45
for my case adding the true flag solved it. I was just passing the data without the true parameter but I wasn't getting the correct response. Thank you. — stanley mbote, Sep 07 '21 at 11:14

score 29 · Answer 2 · answered Jul 05 '13 at 12:13

To convert a valid JSON string back, you can use the json_decode() method.

To convert it back to an object use this method:

$jObj = json_decode($jsonString);

And to convert it to a associative array, set the second parameter to true:

$jArr = json_decode($jsonString, true);

By the way to convert your mentioned string back to either of those, you should have a valid JSON string. To achieve it, you should do the following:

In the Coords array, remove the two " (double quote marks) from the start and end of the object.
The objects in an array are comma seprated (,), so add commas between the objects in the Coords array..

And you will have a valid JSON String..

Here is your JSON String I converted to a valid one: http://pastebin.com/R16NVerw

Yes! Instead of typecasting into an array, we can use the _assoc_ parameter of `json_encode`. Quoting the [PHP docs](http://php.net/manual/en/function.json-decode.php): "When `TRUE`, returned objects will be converted into associative arrays." — CPHPython, May 25 '18 at 10:00

score 10 · Answer 3 · edited Jul 14 '19 at 15:32

10

you can use this for example

$array = json_decode($string,true)

but validate the Json before. You can validate from http://jsonviewer.stack.hu/

edited Jul 14 '19 at 15:32

rawplutonium

386
5
15

answered Jul 05 '13 at 12:26

Nabeel Arshad

467
2
7
22

score 7 · Answer 4 · answered Jul 08 '22 at 18:59

If you don't pass second parameter, or pass false, json_decode() will parse JSON to stdClass object therefore you can use "->" arrow notation to access the object properties.

<?php

// Store JSON data in a PHP variable
$json = '{"email":"john@doe.com"}';

$obj = json_decode($json, false);
print $obj->email;

?>

Output

john@doe.com

score 0 · Answer 5 · answered Nov 29 '22 at 22:04

@Miro Markaravanes save me tons of time of this. In my case I stored the string of a JSON object similar to OP's in my database. When calling it back it had double quotes at the beginning and end of the string.

Due to this every time I tried to using json_decode it wouldn't work.

So I did the following:

    $data = substr($data, 1);
    $data = substr($data, 0, -1);
    $data = json_decode($data);

score -1 · Answer 6 · edited Dec 06 '18 at 15:53

-1

Try with json_encode().

And look again Valid JSON

edited Dec 06 '18 at 15:53

Balaji Rajendran

357
5
17

answered Jul 05 '13 at 12:01

drag.spas

59
6

How to convert a string to JSON object in PHP

6 Answers6

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