php – 如何通过form_open_multipart将id从视图发送到Controller CodeIgniter

我正在尝试从视图发送id到CodeIgniter中的控制器.我的要求是根据按钮id切换功能.这是我的 HTML代码.

查看HTML

<?PHP echo form_open_multipart('upload_control/switch_load','id="bt_addImage"');?>
 <input id= "bt_addImage" type="submit" value="Add Image" /> <br>
 <?PHP echo form_open_multipart('upload_control/switch_load','id="bt_chooseImage"');?>
 <input type="submit" id="bt_chooseImage" value="Submit"/><br>

Upload_control.PHP代码

public function switch_load($id)
{
    if($id == "bt_addImage")
    {
        do_loadcategories();
    }
    else
    {
        do_upload();
    }
}
public function do_loadcategories()
{
    //code list categories
}
public function do_upload()
{
    //code to upload
}

这是正确的吗？

要么

还有其他办法吗？

帮我解决问题.

解决方法

在视图中

$hiddenFields = array('id' => 'bt_addImage'); # add Hidden parameters like this
echo form_open_multipart('upload_control/switch_load','',$hiddenFields);

在控制器中

public function switch_load()
{
    $id = $this->input->post('id');
    if($id == "bt_addImage")
    {
        do_loadcategories();
    }
    else
    {
        do_upload();
    }
}

看起来像这样

<form method="post" accept-charset="utf-8" action="http:/example.com/index.PHP/upload_control/switch_load">
<input type="hidden" name="id" value="bt_addImage" /> # hidden filed.

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