codeigniter返回對象

Question

我從數據庫中選擇數據併成功返回數據。我只是不知道如何訪問陣列的元素，如用戶名或名稱等。這是輸出：

Array（[0] => stdClass Object（[id] => 9 [fullname] =>خاÙ「د اÙ」غاÙ...دي[會員] =>免費[username] => kkkkk [密碼] => 16d7a4fca7442dda3ad93c9a726597e4 [email] => someone @ gmail .com [about] => [city] => riyadh [profilepic] => /home2/sdds/public_html/uploads/Red_velvet_cupcakes_with_roses.jpgRed_velvet_cupcakes_with_roses.jpg [mobile] => [telephone] => [address] => gallery_link] =>））

，這是在模型返回上面的代碼：

public function login() 
{ 
    $this->db->select('*'); 
    $this->db->from('members'); 
    $this->db->where('username', $this->input->post('username')); 
    $this->db->where('password', md5($this->input->post('password'))); 
    $query = $this->db->get(); 
    return $query->result(); 
} 

所以在這裏做什麼？在此先感謝

Answer 1

請檢查：

//model function in your model  
public function login() 
    { 
     $this->db->select('*'); 
     $this->db->from('members'); 
     $this->db->where('username', $this->input->post('username')); 
     $this->db->where('password', md5($this->input->post('password'))); 
     $query = $this->db->get(); 
     return $query->row(); 
    } 
    // in controller 
    $logindata = $this->YOUR_MODEL_NAME->login(); 
// access like this 

    echo $logindata->fullname; 

編號：http://ellislab.com/codeigniter/user-guide/database/results.html

Answer 2

打樁 - 也可以肯定，並檢查如果你真的在你的模型方法有什麼

// check how many records returned with num_rows() 
if ($query->num_rows() == 1) { 

     // since there is just one record, use row() 
     // when 2 or more records, use result() 
     return $query->row(); 
    } 
    else 
    { return false; } 

這樣在你的控制器中，你立即知道是否有任何東西回來 ，你可以顯示相應的視圖。

if(! $logindata = $this->login_model->login()) { 
    // NO login data came back 
    // show appropriate view 

    else { 
// pass to data if you need it in the view 
    $data['logindata'] = $logindata ; 
    // call and show the view 

}