我想用multi-select更新mysql表。选项值在mysql中。我的问题是，多选只显示一个公司，即使员工有多个公司。
例如：

employee1有3个公司，但在select中只显示1。我不能更新这个。但是如果一个员工有一个公司，我改变了它，然后我更新了它，它就工作了。我怎样才能在“全选公司”中显示员工拥有的，而不仅仅是1个？
我的表结构：

索引.php

<form method="post" id="insert_form">  
   <label>Employee name:</label>  
   <input type="text" name="name" id="name" class="form-control" />  
   <br />  
   <label>Address:</label>  
   <textarea name="address" id="address" class="form-control"></textarea>  
   <br />  
   <label>Company:</label>
   <select name="company" id="company" class="form-control" multiple>
      <?php 
         $query2 = "SELECT * FROM company GROUP BY company_id";  
         $result2 = mysqli_query($connect, $query2);  
         while($row2= mysqli_fetch_array($result2)){
      ?>
      <option value="<?php echo $row2['company_id'];?>"><?php echo $row2['name'];?></option>
      <?php
         }
      ?>
   </select>
   <br/>

   <input type="hidden" name="employee_id" id="employee_id" />  
   <input type="submit" name="insert" id="insert" value="Insert" class="btn btn-success" />  
</form>  

<script>
   $(document).on('click', '.edit_data', function(){  
           var employee_id = $(this).attr("id");  
           $.ajax({  
                url:"fetch.php",  
                method:"POST",  
                data:{'employee_id':employee_id},  
                dataType:"json",  
                success:function(data){  
                     $('#name').val(data.name);   
                     $('#address').val(data.address);
                     $('#company').val(data.company);

                     $('#employee_id').val(data.id);  
                     $('#insert').val("Update");  
                     $('#add_data_Modal').modal('show');  
                }  
           });  
      });  
</script>

获取.php

if(isset($_POST["employee_id"]))  
 {  
        $query = "SELECT employee.employee_id AS id, employee.name AS name, employee.address AS address, company.company_id AS company
            FROM employee 
            LEFT JOIN employee_company ON employee.employee_id = employee_company.employee_id
            LEFT JOIN company ON company.company_id = employee_company.company_id

            WHERE employee.employee_id = '".$_POST["employee_id"]."'";  
      $result = mysqli_query($connect, $query);  
      $row = mysqli_fetch_array($result);  
      echo json_encode($row);  
 }