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json – Get a PHP object property that is a number

Posted by: admin April 23, 2020 Leave a comment

Questions:

I’m trying to get a property from JSON data decoded into a PHP object. It’s just a YouTube data API request that returns a video object that has a content object liks so;

[content] => stdClass Object
                (
                    [5] => https://www.youtube.com/v/r4ihwfQipfo?version=3&f=videos&app=youtube_gdata
                    [1] => rtsp://v4.cache7.c.youtube.com/CiILENy73wIaGQn6pSL0waGIrxMYDSANFEgGUgZ2aWRlb3MM/0/0/0/video.3gp
                    [6] => rtsp://v6.cache3.c.youtube.com/CiILENy73wIaGQn6pSL0waGIrxMYESARFEgGUgZ2aWRlb3MM/0/0/0/video.3gp
                )

Doing

$object->content->5

Throws “unexpected T_DNUMBER” – which makes perfect sense. But how do I get the value of a property that is a number?

I’m sure I should know this. Thanks in advance.

How to&Answers:

This should work:

$object->content->{'5'}

Answer:

Another possibility is to use the 2nd parameter to json_decode:

$obj = json_decode(str, true);

You get an array instead of a PHP object, which you can then index as usual:

$obj['content'][5]

Answer:

JSON encode, and then decode your object passing true as the second param in the decode function. This will return an associative array.

$array = json_decode(json_encode($object), true);

Now you can use your new array

echo $array['content']['5'];

Using $object->content->{'5'} will not work if the object was created by casting an array to an object.

A more detailed description can be found here: https://stackoverflow.com/a/10333200/58795

Answer:

Another aproach is casting the object to array.

$array = (array) $object;
$array['content'][5];