Question

这是我的场景，我正在创建一个带有codeigniter的表单，我理解如何使用模型等填充字段。我有表格的布局。它现在从我的索引函数运行。我想存储给该表单的所有数据，并在postdata数组中访问它们，每个索引都是值的名称。请帮忙。 CodeIgniter，PHP

Answer 1

您创建表单

echo form_open('mycontroller/mymethod'); 
// rest of form functions

or  <form name="myform" method="post" action="<?php echo site_url('mycontroler/mymethod');?>" >  // rest of html form

then, in Mycontroller:

   function mymethod()
   {
     $postarray = $this->input->post();

    // you can pass it to a model to do the elaboration , see below
    $this->myloadedmodel->elaborate_form($postarray)
   }   

Model:

  function elaborate_form($postarray)
  {
    foreach($postarray as $field)
    {
      \\ do your stuff
    }
  }

如果您想要XSS过滤，可以在$this->input->post()调用中传递TRUE作为第二个参数。查看user guide on input library

Answer 2

请参阅code igniter's input class

如何构建代码的一个例子是：

public function add_something()
  {
    if (strtolower($_SERVER['REQUEST_METHOD']) == 'post') { // if the form was submitted
      $form = $this->input->post(); 
      // <input name="field1" value="value1 ... => $form['field1'] = 'value1'
      // you should do some checks here on the fields to make sure they each fits what you were expecting (validation, filtering ...)

      // deal with your $form array, for exemple insert the content in the database

      if ($it_worked) {
        // redirect to the next page, so that F5/reload won't cause a resubmit
        header('Location: next.php');
        exit; // make sure it stops here
      }

      // something went wrong, add whatever you need to your view so they can display the error status on the form
    }

    // display the form
  }

通过这种方式，您的表单将被显示，如果提交，其内容将被处理，如果发生错误，您将能够保持提交的值在表单中预先输入，显示错误消息等...并且如果它有效，则用户被重定向到他可以安全地重新加载页面而不需要多次提交。

帮助存储带有codeigniter的postdata

2 个答案: