我正在尝试在CodeIgniter中从视图向控制器发送id。我的要求是根据按钮id切换函数。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
}
它正确吗?
或
有别的办法吗?
帮我解决这个问题。
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();
}
}
视图看起来像这个
<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.
Codeigniter Form Helper
的更改
if($id == "bt_addImage")
进入
if($this->input->post('id') == "bt_addImage")