本文介绍了如何从codeignator中的视图调用控制器函数?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
hi每个人都在这里与codeignator! i的控制器如下:
hi everybody m new here with codeignator!! 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"; } } ?>并且视图位于以下位置:
and view is in the following:
<form action="<?php echo base_url();?>myController/myFn" method="post" name="myform"> <input type="submit" name="submit" value="submit"/> </form>问题是,当我运行视图通过转到localhost单击提交m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!!!
**The requested URL /CodeIgniter/myController/myFn was not found on this server.**但是当我放置 ** http:// localhost / CodeIgniter / index .php / myController / myFn ** 我得到正确的视图输出任何人都在这里谁可以帮助我在这方面thnx提前....
but when i put **localhost/CodeIgniter/index.php/myController/myFn** i got the correct output of the view anybody is here who can help me in this regard thnx in advance....
推荐答案查看侧
<form method="post" action="<?php echo base_url();?>index.php?controlerfile/test" class="form"> Message: <input type="text" name="message"> <input type="submit" value="submit"> <div> You have entered: <?php echo $message; ?> </div>controllerfile
controllerfile
function test() { $data['message']= $this->input->post('message'); //$this->output->append_output($message); $this->load->view('backend/test',$data); }希望这会帮助你..
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如何从codeignator中的视图调用控制器函数?
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