使用php DEFINE中的变量的Twitter API错误

Question

我正在尝试将用户名和密码变量传递给twitter凭据,但它会一直返回,我没有通过身份验证.但是,当我使用实际的用户名和密码而不是变量时,它会成功授权.

$username = $_POST["username"];
$password = $_POST["password"];

$url = "http://search.twitter.com/search.atom?q=golf&show_user=true&rpp=100";
$search = file_get_contents($url);

$regex_name = '/\<name\>(.+?) \(/';
preg_match_all($regex_name,$search,$user);
for($i=0;$user[1][$i];$i++)
{
$follow = $user[1][$i];
    define('TWITTER_CREDENTIALS', '$username:$password');
    $url = "http://twitter.com/friendships/create/".$follow.".xml";
    $ch = curl_init();
    curl_setopt($ch, CURLOPT_URL,$url);
    curl_setopt($ch, CURLOPT_POST, 1);
    curl_setopt($ch, CURLOPT_USERPWD, TWITTER_CREDENTIALS);
    $result= curl_exec ($ch);
    curl_close ($ch);

我认为它与用户名和密码之间的冒号有关,或者可能试图在define函数中使用变量.

有线索吗？

Answer 1

$username = $_POST["username"];
$password = $_POST["password"];
// INCORRECT. Will literary assign TWITTER_CREDENTIALS as $username:$password
// define('TWITTER_CREDENTIALS', '$username:$password');

// CORRECT, will parse the variables and assign the result to TWITTER_CREDENTIALS
define('TWITTER_CREDENTIALS', "$username:$password");

请记住字符串中带双引号(")解析变量的字符串,带单引号的字符串(')不记住.

阅读PHP中有关字符串的更多信息

• PHP字符串
• PHP字符串中的变量解析
• PHP字符串函数