如何在Codeigniter中从视图调用控制器功能？ [英] How to call controller function from view in codeigniter?

问题描述

使用codeigniter，我有一个如下所示的控制器：

With codeigniter, I have a controller as in the following:

<?php if(!defined ('BASEPATH')) exit('not found basepath');

class **myController** extends CI_Controller{

    function __constructor(){
        parent::__constructor();
    }
    public function index(){
        $this->load->view('myview');
    }
    **public function myFn()**{
        echo "my controller is called"; 
    }
}

?>

，视图如下：

<form action="<?php echo base_url();?>myController/myFn" method="post" name="myform">
<input type="submit" name="submit" value="submit"/>
</form>

问题是，当我单击提交时，通过转到本地主机运行视图时，提示出现以下错误！

the problem is that when I run the view by going to localhost after clicking at the submit m intimating by the following error!

在此服务器上找不到请求的URL / CodeIgniter / myController / myFn。

The requested URL /CodeIgniter/myController/myFn was not found on this server.

但是当我放 ** http：//localhost/CodeIgniter/index.php/myController/myFn** 我得到了视图的正确输出

but when I put **http://localhost/CodeIgniter/index.php/myController/myFn** I got the correct output of the view

推荐答案

在您的控制器中，只需删除'**'。

In your controller just delete ' ** '.

<?php if(!defined ('BASEPATH')) exit('not found basepath');

class myController extends CI_Controller{

    function __constructor(){
        parent::__constructor();
    }

    public function index(){
        $this->load->view('myview');
    }

    public function myFn(){
    echo "my controller is called"; 
    }
}
?>

在您看来，尝试：

<?php echo form_open('myController/myFn'); ?>
<?php echo form_submit('submit','SUBMIT'); ?>

这篇关于如何在Codeigniter中从视图调用控制器功能？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！