Question

我正在使用codeigniter中的from，其中程序控制正在转移提交函数，我可以通过添加die函数来测试它。尽管set_rules（）成功检查了条目，但控件未传递给if（$ this-＆gt; form_validation-＆gt; run（））此函数。它已经退出并运行die-2，我一直在测试程序流程。

以下是我的控制器代码

function addPost(){
$this->load->library('form_validation')
if($this->admin_lib->checkMembers()){
if($this->input->post('submit')){

   //validate the form
$this->form_validation->set_rules('country','Country','required');
$this->form_validation->set_rules('city','City','required');
$this->form_validation->set_rules('area','Area','required');
$this->form_validation->set_rules('street','Street','required');
$this->form_validation->set_rules('house_no','House number','required|numeric');

if($this->form_validation->run()){
   //add to database
   die("dead-1");
    if($this->members_model->addPost())
    {
    echo "Successfully made one entry will be validated";

    }
    else{
        echo "Error uploading the datas into database Please contact us about the problem";
            }
        }
   die("Dead -2");
    }
    $data['content']=$this->load->view('members/addPost','',true);
    $this->load->view('members/home',$data);

    }
    else{
    echo "you dont have preveledge to access this page ,<br/> LOgin link rakhnu paryo ";
    }
   }

Answer 1

您的代码将始终调用die（＆＃34; Dead -2＆＃34;），因为它不是Else块的一部分。它直接位于您的If语句下方，这意味着，无论您的表单验证发生什么，它都将永远消亡。

考虑将您的代码更改为以下

if($this->form_validation->run())
{
    //add to database
    die("dead-1");
    if($this->members_model->addPost())
    {
        echo "Successfully made one entry will be validated";
    }
    else
    {
        echo "Error uploading the datas into database Please contact us about the problem";
    }
}
else
{
    die("Dead -2");
}

Codeigniter表单验证：运行功能不起作用

1 个答案: