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我编写了一个CodeIgniter应用程序,用于输入Twitter的屏幕名称,然后从Twitter API 1.1中获取有关此用户的数据。如果Twitter的网名不存在,它返回以下错误:验证不存在的Twitter用户
stdClass Object ([errors] => Array ([0] => stdClass Object ([message] => Sorry, that page does not exist [code] => 34)))
我的问题是,我该如何验证在我的功能get_user_data(...)
上述错误/检查?如果页面/用户不存在,我想返回34
,以便我可以显示适当的未找到的错误视图。任何援助将不胜感激。提前致谢!
// ...
$this->form_validation->set_rules('username', 'username', 'required');
$username = $this->input->post('username');
if($this->form_validation->run() === FALSE || $this->val_username($username) === FALSE)
{
$data = 'Validation Problems';
$this->view('search_page2',$data);
}
else
{
$twitter = new TwitterAPIExchange($settings);
$friends_list_url = 'https://api.twitter.com/1.1/users/show.json';
$friends_list = json_decode($this->get_user_data($username, $twitter, $friends_list_url));
}
}
}
private function get_user_data($username, $twitter, $url)
{
$getfield = '?screen_name=' . $username;
$request_method = 'GET';
return $twitter->setGetfield($getfield)
->buildOauth($url, $request_method)
->performRequest();
}
// ...
+1哇,它的工作,谢谢! – Anthony