我在使用OAuth与PHP Codeigniter连接到Google Contacts API时遇到问题。看来我可以连接,但接收
Message: file_get_contents(https://www.google.com/m8/feeds/contacts/default/full?&max-results=50&oauth_token=<token string>): failed to open stream: HTTP request failed! HTTP/1.0 403 Forbidden
在最后的过程中。这是代码,
// Google Project API Credentials
$Google_api_client_id = 'client-id';
$Google_client_secret = 'client-secret';
$Google_redirect_url = base_url() . 'contact_import/';
$Google_contact_max_result = 50; // integer value
$authcode= $_GET['code'];
$clientid=$Google_api_client_id;
$clientsecret=$Google_client_secret;
$redirecturi=$Google_redirect_url ;
$fields=array(
'code'=> urlencode($authcode),
'client_id'=> urlencode($clientid),
'client_secret'=> urlencode($clientsecret),
'redirect_uri'=> urlencode($redirecturi),
'grant_type'=> urlencode('authorization_code')
);
$fields_string="";
foreach($fields as $key=>$value)
{ $fields_string .= $key.'='.$value.'&'; }
$fields_string=rtrim($fields_string,'&');
//open connection
$ch = curl_init();
//set the url, number of POST vars, POST data
curl_setopt($ch,CURLOPT_URL,'https://accounts.google.com/o/oauth2/token');
curl_setopt($ch,CURLOPT_POST,5);
curl_setopt($ch,CURLOPT_POSTFIELDS,$fields_string);
// Set so curl_exec returns the result instead of outputting it.
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
//to trust any ssl certificates
curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false);
//execute post
$result = curl_exec($ch);
//close connection
curl_close($ch);
//extracting access_token from response string
$response = json_decode($result);
$accesstoken= $response->access_token;
if( $accesstoken!="")
$_SESSION['token']= $accesstoken;
//passing accesstoken to obtain contact details
$xmlresponse= file_get_contents('https://www.google.com/m8/feeds/contacts/default/full?&max-results='.$Google_contact_max_result.'&oauth_token='.$_SESSION['token']);
//reading xml using SimpleXML
$xml= new SimpleXMLElement($xmlresponse);
$xml->registerXPathNamespace('gd', 'http://schemas.google.com/g/2008');
$result = $xml->xpath('//gd:email');
$count = 0;
foreach ( $result as $title )
{
$fetched_email = $title->attributes()->address;
$contact_key[] = $this->db->insert_contact_gmail($fetched_email);
}
如果我在浏览器中复制粘贴带令牌的URL,我可以看到XML响应。请帮忙..
答案 0 :(得分:1)
此错误来自allow-url-fopen
命令
您必须在php.ini
设置CREATE TABLE dbo.AgentAndOsFilter
(
Id INT IDENTITY(1, 1),
AgentVersion VARCHAR(8),
Os VARCHAR(16)
)
,如果您在线工作,则必须从网站adminestor上提出
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