从数据库中选择多个中获取存储的值，并在视图中将它们与图像分开显示，php codeigniter

如何解决从数据库中选择多个中获取存储的值，并在视图中将它们与图像分开显示，php codeigniter

我有一个表单，可以在project_add.PHP视图中添加一个项目，并且表单中有一个多选下拉菜单（支持人员），我能够成功地将其数据作为字符串插入数据库（同一行）中，通过使用下面的代码这是project_add.PHP视图

<div class="form-group col-md-4">
              <label for="pwd">Add Support Staff</label>
              <select id="addStaffMulti" placeholder="Selecct" multiple="multiple" name="staff[]" value="<?PHP echo set_value('staff'); ?>">
              <div class="alert-danger"><?PHP echo form_error('staff'); ?></div>
                <option value="">Select Staff</option>
            </select>
            </div>

这是我的project_model.PHP

function add()
 {
    $arr['project_name'] = $this->input->post('Pname');
    $arr['client_name'] = $this->input->post('Cname');
    $arr['company'] = $this->input->post('PassignTo');
    $arr['project_manager'] = $this->input->post('manager');
    $arr['support_staff'] = $this->input->post('staff');
    $value = implode(",",($this->input->post('staff')));
    $arr['support_staff'] = $value;
    $this->db->insert('projects',$arr);
 }

数据库看起来像：

现在，我想在project_index.PHP视图的表中分别显示每个数据，换句话说，对于下拉菜单的每个选定数据，我想在其中添加一个图像（平均值），这表示是否选择了一个选项来显示一幅图像，如果选择两个选项，则显示两幅图像。这是我的project_index.PHP

<tbody>
                <?PHP
                foreach($project as $n)
                {
                    
                    ?>
                    <tr>
                        <td><?PHP echo $n->project_name;?></td>
                        <td><?PHP echo $n->client_name;?></td>
                        <td><?PHP echo $n->company;?></td>
                        <td><a class="pro-circle"><img class="img-sm" src="<?PHP echo site_url('assets/image/man.png');?>"  data-toggle="tooltip" data-placement="bottom" title="<?PHP echo $n->project_manager;?>" data-original-title="Click to deactivate the user"></a>&nbsp&nbsp </td>
                        <td>
                        <a class="pro-circle"><img class="img-sm" src="<?PHP echo site_url('assets/image/man.png');?>"   data-toggle="tooltip" data-placement="bottom" title="<?PHP echo $n->support_staff;?>"  data-original-title="Click to deactivate the user"></a>&nbsp&nbsp </a>   
                         <!-- <a class="pro-circle"><img class="img-sm" src="<?PHP echo site_url('assets/image/man.png');?>" data-toggle="tooltip" data-placement="bottom" title="<?PHP echo $n->support_staff;?>" data-original-title="Click to deactivate the user"></a>&nbsp&nbsp
                        <a class="pro-circle"><img class="img-sm" src="<?PHP echo site_url('assets/image/man.png');?>" data-toggle="tooltip" data-placement="bottom" title="<?PHP echo $n->support_staff;?>" data-original-title="Click to deactivate the user"></a>
                         -->
                        </td>
                        <td><span class="icoact"></span> Active</td>
                        <td><a class="edit" href="<?PHP echo site_url('admin/project/edit/'.$n->id);?>"><i class="fa fa-pencil-square-o" ></i></a>&nbsp&nbsp
                        <a class="delete"  href="<?PHP echo site_url('admin/project/delete/'.$n->id);?>"  onclick="return confirm('Are you sure want to Delete this Record?')"><i class="fa fa-trash-o"></i></a></td>
                    </tr>
                <?PHP
                }
                ?>
                </tbody>

这是project.PHP控制器

public function index ()
{
  $data['company_name'] = $this->project_model->getAllCompanyName();
  $data['project'] = $this->project_model->getProjectDetails();

  $this->load->view('admin/project/index',$data);
}

更清楚一点：这是project_index.PHP视图，现在仅在support_staff列中显示一幅图像，但是根据我的数据库，我希望显示三幅图像：