PHP - Fast way to Convert JSON Object into Variables

Question

I have a JSON object that can be echoed to view in a browser as follows:

stdClass Object
    (
        [webmaster] => "data"
        [analytics] => "data"
        [facebook] => "data"
        [twitter] => "data"
        [maintenance] => 1
    )

data are other values.

I get the above output using:

    $data = json_decode($domainSpecific);
    print_r($data);

what would be a good way to convert this JSON data into 5 variables preferably with the names of the JSON values - $webmaster, $analytics, $facebook, $twitter, $maintenance?

thankyou

You mean dinamically? That's it, if you add another property to the object like... myspace it would have to create automatically a var called $myspace? — Antonio Laguna
– Antonio Laguna, Commented Jan 28, 2013 at 6:45

Karan Punamiya · Accepted Answer · 2013-01-28 06:53:05Z

1

$data = json_decode($domainSpecific);
foreach($data as $key=>$value)
{
    $$key=$value;
}

answered Jan 28, 2013 at 6:53

Karan Punamiya

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Antonio Laguna · Accepted Answer · 2013-01-28 06:55:02Z

1

Though I'm not sure why would you do that and I'm also not sure if this is a good way to program, here you have something that works as I've tried:

class Test {
    public $webmaster, $analytics, $facebook, $twitter, $manteinance;
}

$test = new Test();
$test->webmaster = 'Trololo';

$object_vars = get_object_vars($test);

foreach ($object_vars as $varname => $value) {
    $$varname = $value;
}

echo $webmaster; //Trololo

answered Jan 28, 2013 at 6:55

Antonio Laguna

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Prasanth Bendra · Accepted Answer · 2013-01-28 07:01:46Z

1

$data = extract(json_decode($domainSpecific, true));
print_r($data);

answered Jan 28, 2013 at 7:01

Prasanth Bendra

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ianace · Accepted Answer · 2013-01-28 09:01:38Z

1

try this php extract()

$data = json_decode($domainSpecific, TRUE);

extract($data , EXTR_PREFIX_SAME);

edit

yes you need to do json_decode with true parameter to return as array

edited Jan 28, 2013 at 9:01

answered Jan 28, 2013 at 7:01

ianace

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