如何从codeigniter的视图中调用控制器功能?-php问题

How to call controller function from view in codeigniter?(如何从codeigniter的视图中调用控制器功能?)

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问题描述

使用 codeigniter，我有一个控制器，如下所示:

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"; 
    }
}

?>

查看如下:

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

问题是，当我通过点击提交后转到本地主机运行视图时，提示以下错误！

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.

但是当我输入 **http://localhost/CodeIgniter/index.php/myController/myFn** 时，我得到了正确的视图输出

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

推荐答案

在你的控制器中删除**".

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'); ?>

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