Twitter的API給錯誤

使用abraham/twitteroauth給出了錯誤

其實，Twitter的窗口來詢問許可時，我接受給出的oauth_token和祕密密鑰，但使用的是當我嘗試發佈獲取此錯誤。

我的代碼如下

2檔

<?php 
session_start(); 
require_once 'classes/twitteroauth.php'; 
define("CONSUMER_KEY", "XXXXXXXXXXXXXXXX"); 
define("CONSUMER_SECRET", "XXXXXXXXXXXXXXXXXXXXXXX"); 
if(isset($_SESSION["oauth_token"])){ 
    $ction = new TwitterOAuth(CONSUMER_KEY, CONSUMER_SECRET, 
       $_SESSION['oauth_token'], $_SESSION['oauth_token_secret']); 
    $tweetmsg = 'Hello World, I am tweeting from my own twitter app!'; 
    $result = $ction->post('statuses/update', array('status' => $tweetmsg)); 
    $httpCode = $ction->http_code; 
    if ($httpCode == 200) { 
     $resultmsg = 'Tweet Posted: '.$tweetmsg; 
    } 
else { 
$resultmsg = 'Could not post Tweet. Error: '.$httpCode.' Reason: '.$result->error; 
}

} 其他{ 回聲「有些該死的錯誤！」給定; } ？>

第一頁

<html> 
<head> 
    <title>Twitter OAuth via popup</title> 
</head> 
<body> 
    <script src="jsjquery.min.js"></script> 
    <script src="js/jquery.oauthpopup.js"></script> 
    <script> 
    $(document).ready(function(){ 
     $('#connect').click(function(){ 
      $.oauthpopup({ 
       path: 'twitter.php', 
       callback: function(){ 
       alert("successfully Tweetted"); 
       window.location.reload(); 
       } 
      }); 
     }); 
     }); 
    </script> 
    <div> 
    <?php 
    echo $resultmsg; 
    ?> 
    </div> 
    <input type="button" value="Connect with Twitter" id="connect" /><br /> 
    <a href="signout.php">Sign Out</a> 
    </body> 
    </html>

第二頁

<?php 
session_start(); 
require_once 'classes/twitteroauth.php'; 
if(!isset($_SESSION["oauth_token"])){ 
define("CONSUMER_KEY", "XXXXXXXXXXXXXXXX"); 
define("CONSUMER_SECRET", "XXXXXXXXXXXXXXXXXXXXXXXXXXXX"); 
$connection = new TwitterOAuth(CONSUMER_KEY, CONSUMER_SECRET); 
$request_token = $connection->getRequestToken(); 
$_SESSION['oauth_token'] = $request_token['oauth_token']; 
$_SESSION['oauth_token_secret'] = $request_token['oauth_token_secret']; 
$url = $connection->getAuthorizeURL($request_token); 
header('Location:'.$url); 
die(); 
} 
?> 
<html> 
<head> 
<title>Share On Twitter</title> 
</head> 
<body> 
<script type="text/javascript"> 
window.close(); 
</script> 
</body> 
</html>

來源

2011-02-02 Harish

的HTML部分，你錯過了一個OAuth的請求，這是交流與請求令牌的第三部分訪問令牌爲用戶帳戶。在documentation of twitteroauth中檢查＃7

<?php 

session_start(); 
require_once 'classes/twitteroauth.php'; 
define("CONSUMER_KEY", "XXXXXXXXXXXXXXXX"); 
define("CONSUMER_SECRET", "XXXXXXXXXXXXXXXXXXXXXXX"); 

if(!$_SESSION['loggedin'] and $_SESSION['oauth_token']) { 
    # Exchange request token with access token 
    $connection = new TwitterOAuth(CONSUMER_KEY, CONSUMER_SECRET, $_SESSION['oauth_token'], $_SESSION['oauth_token_secret']); 
    $token_credentials = $connection->getAccessToken($_REQUEST['oauth_verifier']); 
    $_SESSION['oauth_token'] = $token_credentials['oauth_token']; 
    $_SESSION['oauth_token_secret'] = $token_credentials['oauth_token_secret']; 
    $_SESSION['loggedin'] = true; 

} 
if($_SESSION['loggedin']) { 
    $ction = new TwitterOAuth(CONSUMER_KEY, CONSUMER_SECRET, $_SESSION['oauth_token'], $_SESSION['oauth_token_secret']); 
    $tweetmsg = 'Hello World, I am tweeting from my own twitter app!'; 
    $result = $ction->post('statuses/update', array('status' => $tweetmsg)); 
    $httpCode = $ction->http_code; 
    if ($httpCode == 200) { 
     $resultmsg = 'Tweet Posted: '.$tweetmsg; 
    } 
}

用上面的代碼替換第一個文件。

來源

2011-02-02 07:46:31 Arvin

1懷疑我將如何在第一個函數中獲得$ _SESSION ['oauth_token']，$ _SESSION ['oauth_token_secret']？ – Harish 2011-02-02 08:27:27

Twitter的API給錯誤

回答

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