我想在更改主下拉菜单时构建更新表单然后更新第二个下拉列表

Posted 2023-03-05

【中文标题】我想在更改主下拉菜单时构建更新表单然后更新第二个下拉列表

【英文标题】：I Want To Build Update Form When I Change Main Dropdown Menu Then update the second dropdown list

【发布时间】：2015-09-03 20:35:58

【问题描述】：

                     <div>
                        <label for="form-field-select-3">State : </label>
                        <br />
                        <select class="chosen-select form-control" id="State" name="State" data-placeholder="Choose a State...">
                            <option value="0"> </option>
                            <?php
                             $query=$this->db->query("select Address_State from ref_address_state  where Address_State_ID='".$property->State_ID."'");
                             $query=$query->row();   
                             foreach($Address_State as $rows)
                             
                                if($rows['Address_State']==$query->Address_State)
                                   

                                            echo "<option value='".$rows['Address_State_ID']."' selected>".$rows['Address_State']."</option>";                                                                    
                                  
                            else
                                  
                                           echo "<option value='".$rows['Address_State_ID']."'>".$rows['Address_State']."</option>";                                                                    

                                  
                                   
                             ?>
                        </select>
                    </div>
                     <div>
                        <label for="form-field-select-3">City : </label>
                        <br />
                        <select class="chosen-select form-control" id="City" name="City" data-placeholder="Choose a City...">
                            <option value="0">  </option>
                           <?php
                           $query=$this->db->query("select city_name from ref_city  where city_id='".$property->City."'");
                             $query=$query->row();   
                             foreach($city_name as $rows)
                             
                                if($rows['city_name']==$query->city_name)
                                   

                                     echo "<option value='".$rows['city_id']."' selected>".$rows['city_name']."</option>";                                                                    
                                    
                                    else
                                    
                                    echo "<option value='".$rows['city_id']."'>".$rows['city_name']."</option>";                                                                    

                                    
                             
                                  ?>

                        </select>
                    </div>
                    <div>
                        <label for="form-field-select-3">Locality : </label>
                        <br />
                        <select class="chosen-select form-control" id="Locality" name="Locality" data-placeholder="Choose a Location...">
                            <option value="0">  </option>
                               <?php
                             $query=$this->db->query("select locality_name from ref_locality  where locality_id='".$property->Locality."'");
                             $query=$query->row();   
                             foreach($locality_name as $rows)
                             
                                if($rows['locality_name']==$query->locality_name)
                                   

                                     echo "<option value='".$rows['locality_id']."' selected>".$rows['locality_name']."</option>";                                                                    
                                    
                                    else
                                    
                                    echo "<option value='".$rows['locality_id']."'>".$rows['locality_name']."</option>";                                                                    

                                    
                             
                                  ?>
                        </select>
                    </div>
                    </div>
                </div>
            </div>
    </div>

【问题讨论】：

请详细说明您希望我们做什么

【参考方案1】：

我想你想要这样的东西

http://www.91weblessons.com/codeigniter-ajax-country-state-city-drop-down/

javascript：

 <script>
    function selectState(country_id)//on change of country select box
    
      if(country_id!="-1")
        loadData('state',country_id);
        $("#city_dropdown").html("<option value='-1'>Select city</option>");
      else
        $("#state_dropdown").html("<option value='-1'>Select state</option>");
        $("#city_dropdown").html("<option value='-1'>Select city</option>");
      
    

    function selectCity(state_id)//on change of state select box
    
      if(state_id!="-1")
       loadData('city',state_id);
      else
       $("#city_dropdown").html("<option value='-1'>Select city</option>");
      
    


    function loadData(loadType,loadId)
      var dataString = 'loadType='+ loadType +'&loadId='+ loadId;
      $("#"+loadType+"_loader").show();
      $("#"+loadType+"_loader").fadeIn(400).html('Please wait... <img src="image/loading.gif" />');
      $.ajax(
        type: "POST",
        url: "loadData",
        data: dataString,
        cache: false,
        success: function(result)
          $("#"+loadType+"_loader").hide();
          $("#"+loadType+"_dropdown").html("<option value='-1'>Select "+loadType+"</option>");
          $("#"+loadType+"_dropdown").append(result);
        
     );
    

    </script>

控制器：

<?php

    public function index()
    
       $this->load->model('model');
       $result['list']=$this->model->getCountry();
       $this->load->view('top');
       $this->load->view('index',$result);
       $this->load->view('footer');
    

     public function loadData()
     
       $loadType=$_POST['loadType'];
       $loadId=$_POST['loadId'];

       $this->load->model('model');
       $result=$this->model->getData($loadType,$loadId);
       $HTML="";

       if($result->num_rows() > 0)
         foreach($result->result() as $list)
           $HTML.="<option value='".$list->id."'>".$list->name."</option>";
         
       
       echo $HTML;
     
    ?>

型号：

<?php
function getCountry()

   $this->db->select('id,country_name');
   $this->db->from('country');
   $this->db->order_by('country_name', 'asc');
   $query=$this->db->get();
   return $query;


function getData($loadType,$loadId)

   if($loadType=="state")
    $fieldList='id,state_name as name';
    $table='state';
    $fieldName='country_id';
    $orderByField='state_name';
   else
    $fieldList='id,city_name as name';
    $table='city';
    $fieldName='state_id';
    $orderByField='city_name';
   
   $this->db->select($fieldList);
   $this->db->from($table);
   $this->db->where($fieldName, $loadId);
   $this->db->order_by($orderByField, 'asc');
   $query=$this->db->get();
   return $query;
 

?>

【讨论】：

请在您的帖子中添加解决方案，而不仅仅是链接，以防链接失效。

@Dakkaron 实际上我很清楚我只放链接但现在已经编辑的问题

每当您使用链接回答问题时，请始终在回答中包含解决方案的关键部分。您的答案可能在此页面上存在多年，但链接可能不在。因此，有相同问题的其他人可能会找到您的帖子，但由于链接已损坏，因此无法对答案做任何事情。感谢您提供答案，我现在会投赞成票而不是投反对票。