无法在codeigniter中通过ajax显示带有php代码的div

无法在codeigniter中通过ajax显示带有php代码的div,php,jquery,html,ajax,codeigniter,Php,Jquery,Html,Ajax,Codeigniter,我是ajax新手。当通过ajax从下拉列表中选择区域时,我试图显示特定的餐厅数据,但我无法正确显示餐厅视图。请指导我怎么做 请指导我如何在控制器中嵌入视图代码 其视图代码: <div class="container"> <table align="centre" class="table table-condensed table-striped table-hover no-margin"style="width:70%">

我是ajax新手。当通过ajax从下拉列表中选择区域时,我试图显示特定的餐厅数据,但我无法正确显示餐厅视图。请指导我怎么做

请指导我如何在控制器中嵌入视图代码

其视图代码:

     <div class="container">


            <table align="centre" class="table table-condensed table-striped table-hover no-margin"style="width:70%">
           <thead>
           <tr style="width: 56%;">
               <th>
                   No.    
               </th>
               <th style="">
                Restaurant Names
               </th>
           </tr>

           </thead>
            <tbody>
            <?php $i=1;foreach($result as $row){
                ?>
            <tr id="<?php echo $row->restaurant_id; ?>" class="res_id">
                <th style="">
                    <?php echo  $i++; ?>
                </th>
                <th style="">
                    <?php echo $row->restaurant_name; ?>

                </th>

                <th style="width: 1%" >
                    <a href="<?php echo base_url();?>index.php/BulkRecipe_Controller/bulk_recipe/<?php echo $row->restaurant_id;?>"  class="btn btn-warning" <i class="glyphicon-edit"></i>See Menu</a>

                </th>
                 </tr>
            <?php } ?>

            </tbody>
            </table>
</div>
Ajax代码:

    function get_rests(){

    var city_id = $('#city_id').val();
    var area_id = $("#area_id").val();
    $.getJSON("<?= base_url();?>index.php/Bulk_Controller/get_rests/"+city_id+"/"+area_id,function(data) {
        console.log(data);
        $.each(data, function(i, j){

            $('#restAgainst').append("<div>"+j.restaurant_name+"</div>");
        });

    });
}
查看代码:

        <div id="restAgainst">

    </div>

尝试以下代码:

视图:(在javascript内部)

型号:

public function name()
{
$this->db->select('*');
$this->db->from('tname');
    $query = $this->db->get();
    $result = $query->result_array();

    return $result;
}
请尝试以下代码:

视图:(在javascript内部)

型号:

public function name()
{
$this->db->select('*');
$this->db->from('tname');
    $query = $this->db->get();
    $result = $query->result_array();

    return $result;
}

无法看到正在调用的ajax函数?一旦选择,控制台上是否有任何结果?请阅读此内容,这是
restAgainst
在同一视图中显示的??两个视图代码应该出现在同一页中。看不到正在调用的ajax函数吗?一旦选择,控制台上是否有任何结果?阅读此信息,这是
restAgainst
出现在同一视图中??两个视图代码应出现在同一页中。
var city_id = $('#city_id').val();
var area_id = $("#area_id").val();
  getRequest("<?= base_url();?>index.php/Bulk_Controller/get_rests/"+city_id+"/"+area_id", function(data) {

    var data = JSON.parse(data.responseText);
    for (var i = 0; i < data.length; i++) {

        $("#name").append("<span>" +data[i].restaurant_name+ "</span>");

    }
});


  function getRequest(url, callback) {
    var request;
    if (window.XMLHttpRequest) {
        request = new XMLHttpRequest(); // IE7+, Firefox, Chrome, Opera, Safari
    } else {
        request = new ActiveXObject("Microsoft.XMLHTTP"); // IE6, IE5
    }
    request.onreadystatechange = function() {
        if (request.readyState == 4 && request.status == 200) {
            callback(request);
            $('.loading').hide();
        }
    }
    request.open("GET", url, true);
    request.send();
}
public function name()
{

    $this->load->model('page_model');

    $name= $this->page_model->name();

    echo json_encode($name);
}   
public function name()
{
$this->db->select('*');
$this->db->from('tname');
    $query = $this->db->get();
    $result = $query->result_array();

    return $result;
}