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我使用 ajax 为我的网站动态加载一些内容，它在加载注册和登录表单等内容时效果很好，因为我不必向视图本身发送任何数据（register_view等）。
但是，当我尝试加载不同的东西时，例如用户的配置文件，它需要我将一些变量传递给视图，并且当我遇到 AJAX 问题时。
而且我确定我发送的变量在控制器中经过测试isset，!empty但是在视图中，它突然变成了未定义的变量，这只发生在通过 AJAX 访问配置文件时。
PHP代码：
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);
                    }
                }
            }

视图（view_Profile_view）： 

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

jQuery/JS 代码： 

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.
        });

}
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1 回答 1

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从外观上看，您分配的 $data['information'] = $query_result;so$information对象可用于视图，但您直接传递$query_result给视图而不是$data

所以改变： $this->load->view("view_Profile_view", $query_result);

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

于 2012-06-12T19:02:31.963 回答