使用PHP連接到Twitpic API

Question

我正在爲應用程序使用Twitpic API。我有點卡住了，因爲我似乎無法找到錯誤。這段代碼我讀過的所有文檔都說這是正確的。幫助將不勝感激。謝謝！

function do_twitpic() 
{   
    $media = 'http://image-to-upload.jpg'; 
    $username = $_POST['username']; 
    $password = $_POST['password']; 

    $postfields = array(); 
    $postfields['username'] = $username; 
    $postfields['password'] = $password; 
    $postfields['media'] = "@".$media; 

    $twitter_url = 'http://twitpic.com/api/upload'; 
    $curl = curl_init(); 
    curl_setopt($curl, CURLOPT_CONNECTTIMEOUT, 2); 
    curl_setopt($curl, CURLOPT_HEADER, false); 
    curl_setopt($curl, CURLOPT_RETURNTRANSFER, 1); 
    curl_setopt($curl, CURLOPT_BINARYTRANSFER, 1); 
    curl_setopt($curl, CURLOPT_URL, $twitter_url); 
    curl_setopt($curl, CURLOPT_POST, 3); 
    curl_setopt($curl, CURLOPT_POSTFIELDS, $postfields); 
    $result = curl_exec($curl); 
    curl_close($curl); 
    $login_xml = new SimpleXMLElement($result); 
    if (isset($login_xml->error)) { 
     print_r($login_xml); 
    } else { 
     print_r($login_xml); 
    } 
} 

非常感謝！

Answer 1

根據此頁面$媒體需要是圖像的二進制編碼數據，而不是它的URL。 http://www.twitpic.com/api.do#upload

Answer 2

@在圖像的「真實路徑」前面，而不是公開的URL，將圖像動態轉換爲二進制數據。