PHP注意：尝试获取非对象的属性

Question

I have following code and I get PHP notices...

$tst_msg = json_decode($received_text); //JSON decode
$user_name = $tst_msg->name; //fetching name
$msg = $tst_msg->message;// fetching messages

I get following notices

PHP Notice: Undefined offset: 1
PHP Notice: Trying to get property of non-object in..

Answer 1

$tst_mst is a array and not an object. For objects you can use the arrow, for arrays you have to adres the key like $test[key_name] .

print_r($tst_msg); gives you the output, then use $test_msg['name'] to get the name (if the key in JSON was 'name').

A tutorial is present on the PHP.net page.