试图在（PHP）中获取非对象的属性

Question

get error message:

Notice: Trying to get property of non-object in C:\\xampp\\htdocs\\test.php on line 11

This is my code:

<?php
$json_url = "http://samp-stats.ru/web/api-12492.js";
$json     = file_get_contents($json_url);
$data     = json_decode($json);

echo "<pre>";
var_dump($json);
echo "</pre>";


for ($i = 0; $i < count($data->plinfo); $i++) {  
    $document->write($data->plinfo[$i]->name + ' - ' + $data->plinfo[$i]->score + '<br>');    
}


?>

Answer 1

The JSON is incorrectly formatted.

• There should be nothing preceding the JSON, just the JSON. So remove the var api = at the beginning.
• All object names and string values need to be in double quotes. So ip:'37.59.30.67' needs to read "ip":"37.59.30.67"
• No trailing commas at the end of a range, eg. {name:'Mariu$Bahaos',score:'48' }, ] . That trailing comma tells the parser to expect another object in the array.
• No semi-colon at the end of your string. } is the end character for a JSON string.

Please read up on JSON at JSON.org