2014-02-09 63 views
1

當試圖簡單地得到提及對正確驗證我得到以下JSON一筆賬:亞伯拉罕的twitteroauth的PHP庫中提到的回頭率不良網址

"mentions": { 
    "errors": [ 
     { 
     "message": "Sorry, that page does not exist", 
     "code": 34 
     } 
    ] 
}, 

,我使用登錄併產生這種情況的代碼如下:

private function get_twitter($json) { 

    $_SESSION['access_token']['oauth_token'] = ACCESS_TOKEN; 
    $_SESSION['access_token']['oauth_token_secret'] = ACCESS_TOKEN_SECRET; 

    /* If access tokens are not available redirect to connect page. */ 
    if (empty($_SESSION['access_token']) || empty($_SESSION['access_token']['oauth_token']) || empty($_SESSION['access_token']['oauth_token_secret'])) { 
     /* Load and clear sessions */ 
     session_start(); 
     session_destroy(); 
     if (CONSUMER_KEY === '' || CONSUMER_SECRET === '' || CONSUMER_KEY === 'CONSUMER_KEY_HERE' || CONSUMER_SECRET === 'CONSUMER_SECRET_HERE') { 
      $json['twitter_login_prompt'] = 'You need a consumer key and secret. Get one from <a href="https://dev.twitter.com/apps">dev.twitter.com/apps</a>'; 
      exit; 
     } 

     /* Build an image link to start the redirect process. */ 
     $json['twitter_login'] = '<a href="./redirect.php"><img src="./images/lighter.png" alt="Sign in with Twitter"/></a>'; 
    } 
    /* Get user access tokens out of the session. */ 
    $access_token = $_SESSION['access_token']; 

    /* Create a TwitterOauth object with consumer/user tokens. */ 
    $connection = new TwitterOAuth(CONSUMER_KEY, CONSUMER_SECRET, $access_token['oauth_token'], $access_token['oauth_token_secret']); 

     /* If method is set change API call made. Test is called by default. */ 
     $json['twitter_verify'] = $connection->get('account/verify_credentials'); 
     $json['mentions'] = $connection->get('/1/statuses/mentions'); 
     return $json; 
    } 

回答

0

您正在使用的API版本已棄用 https://dev.twitter.com/docs/api/1/get/statuses/mentions

請考慮將使用1.1版本的Twitter的API的: https://dev.twitter.com/docs/api/1.1/get/statuses/mentions_timeline

您可能需要使用下面的指令,而不是:

$json['mentions'] = $connection->get('/1.1/statuses/mentions'); 
+1

非常感謝,我想通了這一點事後。 由於某種原因,我無法使用部分網址。只有完全合格的網址。 – bwavoplz

+0

不客氣@bwavoplz!你會如此善良接受我的回答嗎? –

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