我是Codeignite中的新成員。我有一個帶有「index()」和「view($ id)」函數的「測試」控制器。索引方法轉到「test1.php」視圖。在「test1.php」中,我有一個下拉選項。如果我在這個頁面上提交,將會調用「測試」控制器的「視圖」方法。我的問題是,如何將選項值傳遞給「view」函數,以便將方法的參數設置爲選項值,然後URL將如"http://localhost/test/view/id"
這些id是來自「test1.php」的選項值顯示html選項值codeigniter
測試控制器
class Test extends CI_Controller{
public function indx() {
//some code
$this->load->view('test1.php');
}
public function view($id)
{
//some code, here I use $id which I want to be option value from test1.php
}
test1.php
<?php
$options = array(
'1' => 'One',
'2' => 'Two',
'3' => 'Three',
'4' => 'Four',
);
$js = 'id="shirts" onChange="this.form.submit();"';
echo form_dropdown('shirts', $options, '1', $js);
/* here I want to call echo form_open() as echo form_open("/test/view/[option value]") but I don't know how to do this;*/