多选下拉列表如何使用jquery ajax表单提交（codeigniter）将所选值保存到数据库

我使用jQueryAjax提交表单，我有一个多选下拉列表，我正在选择并将值保存到数据库基本上所有选择的值都保存在数据库中，但当我检索视图中的值时，只有一个值正在提交

<div class="form-group ">
<label class="control-label col-lg-2">Location</label>                
<div class="col-lg-2">
<select class="form-control" id="city"   onChange="getArea(this.value);">
<option value="">- Select Location -</option> 
<?php  foreach($city_tuitor as $rom) { ?>
<option value="<?php echo $rom->city_id; ?>"> <?php echo $rom->city_name; ?>        </option>
<?php  } ?>      
</select>
</div>
<div class="col-lg-2">
<select class="form-control" id="area" name="area" multiple>
<option value="">- Choose Area -</option>
</select>
</div>

这是我试图得到所有面积值的视图，但我只得到面积的第一个值

<?php 
foreach($search_agent as $agent){
?>
<tr>
<td>City:</td>
<td><?php echo $agent-> city_name; ?></td>
</tr>
<tr>
<td>Area</td>
<td>
<?php
echo $agent-> area_name;  ?>                      
</td>
</tr>
<?php } <?PHP

---

您的意思是成功插入数据库，但检索有问题？是的，值插入成功，我可以打印如下数组（[city\u id]=>1[area\u id]=>1,2,3,4//这些是值）表单提交成功基本上我不确定是否正确：（好的，您的查询从哪里检索结果公共函数search\u agent\u view（）{$tutor\u name=$this->session->username；$this->db->join（'city\u tuitor'，'search\u agent\u tuitor.city\u id=city\u tuitor.city\u id'）；$this->db->join（'area\u tuitor'，'search\u agent\u tuitor.area\u id=area\u tuitor.area\id'）；$this db->select（'*）；$this->db->from（'search_-agent'）；$this->db->where（'tutor_-name'，$tutor_-name）；$query=$this->db->get（）；return$search_-agent=$query->result（）}

public function add_searchagent()
{

$city=$_POST['city'];

//if(isset($_POST['area'])) $area=implode(",",$_POST['area']);
//$area=implode(',',$_POST['area']);
$area=$_POST['area'];
$data = array(
   'city_id' => $city,
   'area_id' => $area
);
//print_r ($data);
$this->db->insert('search_agent', $data); 
echo "Form Submitted Succesfully";
}
}

<?php 
foreach($search_agent as $agent){
?>
<tr>
<td>City:</td>
<td><?php echo $agent-> city_name; ?></td>
</tr>
<tr>
<td>Area</td>
<td>
<?php
echo $agent-> area_name;  ?>                      
</td>
</tr>
<?php } <?PHP

public function search_agent_view()
{
$tutor_name = $this->session->username;

$this->db->join('city_tuitor', 'search_agent_tuitor.city_id = city_tuitor.city_id');
$this->db->join('area_tuitor', 'search_agent_tuitor.area_id = area_tuitor.area_id');
$this->db->select('*');
$this->db->from('search_agent_tuitor');
$this->db->where('tutor_name', $tutor_name);
$query=$this->db->get();
return $search_agent = $query->result();

}