问题描述
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>   </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>   </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>  
<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>  
<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:
解决方法
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