[英]Php function variable passing
如何让 function is_twitter_logged_in() 识别在之前的 function 中分配的变量 $Twitter?
function twitter_logged_in($twitter_user)
{
global $consumer_key;
global $consumer_secret;
$Twitter = new EpiTwitter($consumer_key, $consumer_secret);
if(isset($_GET['oauth_token']) || (isset($_COOKIE['oauth_token']) && isset($_COOKIE['oauth_token_secret'])))
{
// user has signed in
if( !isset($_COOKIE['oauth_token']) || !isset($_COOKIE['oauth_token_secret']) )
{
// user comes from twitter
// send token to twitter
$Twitter->setToken($_GET['oauth_token']);
// get secret token
$token = $Twitter->getAccessToken();
// make the cookies for tokens
setcookie('oauth_token', $token->oauth_token);
setcookie('oauth_token_secret', $token->oauth_token_secret);
// pass tokens to EpiTwitter object
$Twitter->setToken($token->oauth_token, $token->oauth_token_secret);
}
else
{
// user switched pages and came back or got here directly, stilled logged in
// pass tokens to EpiTwitter object
$Twitter->setToken($_COOKIE['oauth_token'],$_COOKIE['oauth_token_secret']);
}
}
elseif (isset($_GET['denied'])) {
// user denied access
echo 'You must sign in through twitter first';
}
else {
// user not logged in
echo 'You are not logged in';
}
global $Twitter;
$twitter_user= $Twitter->get_accountVerify_credentials();
// show screen name (not real name)
echo $twitter_user->screen_name;
// show profile image url
//$twitter_image = $user->profile_image_url;
return $Twitter;
}
function is_twitter_logged_in($Twitter) {
global $Twitter;
$twitter_user = $Twitter;
if ( $twitter_user->screen_name == '' ){return false;}
else {return true;}
}
$Twitter
必须首先在 function scope 之外定义,然后可以使用global $Twitter
在函数中引用它,并且is_twitter_logged_in()
甚至不使用$Twitter
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.