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我正在使用这段代码来尝试获取我所有的最新推文以在我的网站上打印，但它返回了一个关于未通过身份验证的错误。我在 Dev 部分创建了一个应用程序并获取了我的使用者和 OAuth 密钥，并将它们添加到代码中的正确位置。

    <?php
        function buildBaseString($baseURI, $method, $params) {
            $r = array();
        ksort($params);
        foreach($params as $key=>$value){
            $r[] = "$key=" . rawurlencode($value);
        }
        return $method."&" . rawurlencode($baseURI) . '&' . rawurlencode(implode('&', $r));
        }

        function buildAuthorizationHeader($oauth) {
        $r = 'Authorization: OAuth ';
        $values = array();
        foreach($oauth as $key=>$value)
            $values[] = "$key=\"" . rawurlencode($value) . "\"";
        $r .= implode(', ', $values);
        return $r;
        }

        $url = "https://api.twitter.com/1.1/statuses/user_timeline.json?screen_name=jameskrawczyk";
        $oauth_access_token = "SECURITY";
        $oauth_access_token_secret = "SECURITY";
        $consumer_key = "SECURITY";
        $consumer_secret = "SECURITY";

        $oauth = array( 'oauth_consumer_key' => $consumer_key,
                    'oauth_nonce' => time(),
                    'oauth_signature_method' => 'HMAC-SHA1',
                    'oauth_token' => $oauth_access_token,
                    'oauth_timestamp' => time(),
                    'oauth_version' => '1.0');



        $base_info = buildBaseString($url, 'GET', $oauth);
        $composite_key = rawurlencode($consumer_secret) . '&' .     rawurlencode($oauth_access_token_secret);
        $oauth_signature = base64_encode(hash_hmac('sha1', $base_info, $composite_key, true));
        $oauth['oauth_signature'] = $oauth_signature;



        $header = array(buildAuthorizationHeader($oauth), 'Expect:');
        $options = array( CURLOPT_HTTPHEADER => $header,
                      //CURLOPT_POSTFIELDS => $postfields,
                      CURLOPT_HEADER => false,
                      CURLOPT_URL => $url,
                      CURLOPT_RETURNTRANSFER => true,
                      CURLOPT_SSL_VERIFYPEER => false);



        $feed = curl_init();
        curl_setopt_array($feed, $options);
        $json = curl_exec($feed);
        curl_close($feed);
        $twitter_data = json_decode($json, true);
        foreach ($twitter_data as $elem)
        {
          print_r($elem);
          echo '<br>';
        }

页面返回错误

Array ( [0] => Array ( [message] => Could not authenticate you [code] => 32 ) )
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2 回答 2

2

答案似乎是那行

$url = "https://api.twitter.com/1.1/statuses/user_timeline.json?screen_name=jameskrawczyk";

最后不应该有?screen_name=jameskrawczyk。

于 2012-11-06T10:36:41.050 回答
1

您必须使用库从 twitter 获取数据
“themattharris”是一个很好的著名库，从这里下载

于 2012-11-06T10:36:55.070 回答