當通過AJAX加載視圖時未定義的變量

Question

我使用ajax爲我的網站動態加載一些內容，加載諸如註冊和登錄表單之類的東西時運行良好，因爲我不必向視圖本身發送任何數據（register_view等）。
但是，當我嘗試加載不同的東西時，例如用戶的個人資料，它需要我將一些變量傳遞給視圖，以及當我遇到AJAX問題時。
我確定我發送的變量在控制器中被isset和!empty測試， 但是在視圖中，它突然變成了未定義的變量，這隻有在通過AJAX訪問配置文件時纔會發生。
PHP代碼：
控制器：

if($this->uri->segment(4)){//if viewing a specific profile. 
       /*escape the uri segment*/ 
       $segment = intval($this->uri->segment(4)); 
       if($segment == 0){//the uri segment was a string 
        /*display error message.*/ 
        $data['content'] = 'redirect_message'; 
        $data['information'] = 'Could\'nt find the profile!, please try again.'; 
        $this->load->view('templates/manage', $data); 
       } 
       else{//else , the uri segment is a number, considered safer. 
        $query_result = $this->db_model->getProfile($segment);//get the Profile 
        /*check if any results were returned.*/ 
        if($query_result->num_rows() > 0){ 
         /*load a view to display the specified Profile.*/ 
         $data['information'] = $query_result; 

         if($this->input->is_ajax_request())//requesting via ajax, display the content only. 
         $this->load->view("view_Profile_view", $query_result); 
         else{ 
          $data['content'] = 'view_Profile_view'; 
          $this->load->view('templates/manage', $data); 

         } 
        } 
        else{ //no rows returned. 
         /*show error message.*/ 
         $data['content'] = 'redirect_message'; 
         $data['information'] = 'Error viewing the Profile!'; 
         $this->load->view('templates/manage',$data); 
        } 
       } 
      } 

視圖（view_Profile_view）：

/*display the profile:*/ 
$row = $information->row();//error occurs here! 
echo $row->username.'</br>'; 
echo $row->email; 

的jQuery/JS代碼：

var base_url = "/"; 
var site_url = "/index.php/"; 



$(document).ready(function(){ 
$('.ajax_anchor').click(function(){ 
    loadForm(this); 
    return false; 
}); 


}); 
function loadForm(anchor){ 
    var splitted_url = $(anchor).attr('href').split("/"); 

    if(splitted_url.length == 7){//probably accessing /site/login or /site/register not something like /site/profiles/view/[ID]. 
     var url = splitted_url[splitted_url.length-2]+"/"+splitted_url[splitted_url.length-1]; 
     } 
    else if(splitted_url.length == 9) {//probabbly accessing /site/profiles/view/[ID] not something like /site/login. 
     var url= 
      splitted_url[splitted_url.length-4]+"/" 
      + 
      splitted_url[splitted_url.length-3]+"/" 
      + 
      splitted_url[splitted_url.length-2]+"/" 
      + 
      splitted_url[splitted_url.length-1]+"/" 

      ; 
    } 

    var csrf_token = $.cookie('csrf_cookie_name');//holding the csrf cookie generated by CodeIgniter, using jQuery cookie plugin 
    $.ajax({ 
      type: "POST", 
      url: site_url+url, 
      data: {csrf_test_name:csrf_token}//pass the csrf token otherwise codeigniter will return an error(500). 
     }).done(function(html) { 
      var ajaxResult$ = $('#ajax_result');//ajax_result is an empty div, used to display ajax results. 
      ajaxResult$.empty().append(html).dialog();//dialog:is a jquery-ui function. 
     }); 

} 

Answer 1

通過它的外觀的分配$data['information'] = $query_result;所以$information對象可用於視圖，而是直接傳遞給$query_result到視圖，而不是$data

因此改變： $this->load->view("view_Profile_view", $query_result);

要： $this->load->view("view_Profile_view", $data);