的file_get_contents忽略錯誤

Question

$url = 'xxxxxxxxx'; 
$result = file_get_contents($url); 
$stuff = json_decode($result, true); 
if (false !== ($contents = $stuff)) { 
echo $contents; 
} else { 
//do nothing 
} 

如果查詢犯規返回任何結果，我想沒有什麼表現

相反，我得到

Warning: file_get_contents(xxxx) [function.file-get-contents]: 
failed to open stream: HTTP request failed! HTTP/1.0 400 Bad Request in /home/xxxxx/xxx.php on line 4 

Answer 1

函數之前添加@或設置error_reporting(0)（不是因爲所有推薦錯誤將被抑制）。