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Codeigniter:没有在控制台上获得第二次结果,但Ajax并没有失败

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我正在处理类似于选择的问题: Country->state->city (通过ajax)在Codeigniter中,我想在下拉列表中加载 Series->season_num->episodes ,使用ajax我得到 season_num 的结果选择 series 但我的ajax或查询(不要't know which) is not running right i'获取空记录,当它将数据返回到ajax为 success 时,我甚至没有在控制台中打印它 .
这是我的Ajax功能:

<script>
 function selectSeasons(series_id)
  {
  // alert(series_id);
   if(series_id != "-1")
   {
     loadData('season',series_id);
     $("#city_dropdown").html("<option value='-1'>Select Episode</option>");  
   }

   else
   {
    $("#season_dropdown").html("<option value='-1'>Select Season Num</option>");
    $("#episode_dropdown").html("<option value='-1'>Select Episode</option>");    
    }
   }

function selectEpisodes(season_id)
 {
  if(season_id != "-1"){
  loadData('episode',season_id);
 }
 else
 {
  $("#episode_dropdown").html("<option value='-1'>Select Episode</option>");    
 }
}

function loadData(loadType,loadId)
{
// season as type and option as its id
var dataString = 'loadType='+ loadType +'&loadId='+ loadId;
// alert(dataString);
$("#"+loadType+"_loader").show();
  $("#"+loadType+"_loader").fadeIn(400).html('Please wait... ');
$.ajax({
  type: "POST",
  url: "<?php echo base_url() ?>seasons/loadData",
  data: dataString,
  cache: false,
  success: function(result)
  {
    console.log('success');
    console.log(result);
    $("#"+loadType+"_loader").hide();
    $("#"+loadType+"_dropdown").html("<option value='-1'>Select "+loadType+"</option>");  
    $("#"+loadType+"_dropdown").append(result);  
  },
   fail: function(result)
   {
    console.log('Fail');
    console.log(result);

  }
});

}

这是我的控制器功能: Season/loadData

public function loadData()
{
    $session_id = $this->session->userdata('id');
    if (isset($session_id)) 
    {
        $loadType = $_POST['loadType'];
        $loadId   = $_POST['loadId'];
        $HTML="";

        if ($loadType = "season") 
        {
            $result = $this->model_season->getSeasonData($loadType,$loadId);
            if($result->num_rows() > 0)
            {
                // print_r($result->result());
                // die();
                foreach($result->result() as $list)
                {
                    $HTML.="<option value='".$list->season_num_id."'>".$list->season_num."</option>";
                }

            }

            echo $HTML;
        }

        else
        {
            $result = $this->model_season->getEpisodeData($loadType,$loadId);
            // echo $result;

            if($result->num_rows() > 0)
            {

                foreach($result->result() as $list)
                {
                    $HTML.="<option value='".$list->episodes_id."'>".$list->episode."</option>";
                }
            }
            echo $HTML;
        }

    }
    else
    {
        redirect('user_authentication');
    }
}

这是我的模型类,我从db获取结果:

function getSeasonData($loadType,$loadId)
{
    $fieldList    ='season_num_id,season_num';
    $table        ='season_num';
    $fieldName    ='seasons_id_fk';
    $orderByField ='season_num';

    $this->db->select($fieldList);
    $this->db->from($table);
    $this->db->where($fieldName, $loadId);
    $this->db->order_by($orderByField, 'asc');
    $query=$this->db->get();
    if ($query) 
    {
        return $query; 
    }

    else
    {
        return FALSE;
    }


}

function getEpisodeData($loadType,$loadId)
{
    $fieldList     ='episodes_id, episode';
    $table         ='episodes';
    $fieldName     ='season_num_id_fk';
    $orderByField  ='episode';  

    $this->db->select($fieldList);
    $this->db->from($table);
    $this->db->where($fieldName, $loadId);
    $this->db->order_by($orderByField, 'asc');
    $query=$this->db->get();

    if ($query->row() > 0) 
    {
        return $query;
    }

    else
    {
        return FALSE;
    }

}

第一次选择 Series 我获得选项,例如1,2,3,4但是当我选择这些数字以获得更多剧集时,它只返回它没有失败的ajax请求,它甚至不打印 Else 条件中的任何内容 . 请帮忙吗?如果需要任何其他文件,请告诉我 . 谢谢

1 回答

  • 0

    在ajax调用中使用错误:而不是失败:获取是否有任何错误

    fail: function(result)
    {
        console.log('Fail');
        console.log(result);
        $("#"+loadType+"_loader").hide();
        $("#"+loadType+"_dropdown").html("<option value='-1'>Select "+loadType+"</option>");  
        $("#"+loadType+"_dropdown").append(result);  
    },
    
    error: function(result)
    {
        console.log(result);
    }
    

    在Codeigniter Note : 如果php代码有任何错误,那么它将返回500状态作为响应,因此ajax success函数无法处理此响应并将其赋予错误函数 . 但正如您使用的代码失败那样,这就是为什么你没有得到任何日志 .

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