笨訪問結果集

Question

我是新來的笨並有下面的代碼：

$userAuthenticated = $this->Membership_model->userLogin($email, $pass); 

和模型，下面是方法：

function userLogin($email, $password) 
    { 
     $this->db->select('*'); 
     $this->db->from('members'); 
     $this->db->where('email', $email); 
     $this->db->where('password', $password); 
     $query = $this->db->get(); 
     return $query->result(); 
    } 

我面臨的訪問問題上面顯示的$ userAuthenticated。它包含嘗試登錄的用戶的詳細信息。我如何訪問屬性？

問候，

Answer 1

取代這個

return $query->result(); 

與此：

if ($query->num_rows() == 1) { 
    return $query->row(); } 
    else { return FALSE; } 

替換此

$userAuthenticated = $this->Membership_model->userLogin($email, $pass); 

與此

// check negative condition first 
    if(! $data['userAuthenticated'] = $this->Membership_model->userLogin($email, $pass)){ 
    // show no results view 
    $this->load->view('wedontknowyouatall', $data); 
    } 
    else { 
    // we have a result back 
    // and we have already passed userAuthenticated into $data 
    // now show result view 
    $this->load->view('mysweetviewpage', $data); 
} 

在視圖頁面，您現在可以訪問所有的數據庫字段，如

echo $userAuthenticated->email ; 

Answer 2

return $query->row(); 

if($userAuthenticated){ 
login true 
}else{ 
login false 
} 

Answer 3

我會檢查由這樣的查詢選擇的行數：獲得選擇

if($userAuthenticated->num_rows() > 0){ 
    login 
}else{ 
    not login 
} 

否則你必須做循環的數據：

foreach($userAuthenticated as $user){ 
    echo $user->[variable]; 
} 

Answer 4

只需使用

$ userAuthenticated-> firstname; $ userAuthenticated-> email; $ userAuthenticated-> pass;

等等。