Php 将值从控制器传递到视图

我创建了一个表单，并从表单中传递了name和picture的值。通过上传控制器访问该值，如下所示：

  $data = array(
    'title' => $this->input->post('title', true),
    'name' => $this->input->post('name',true),
    'picture' => $this->file_upload($_FILES['picture'])
);
return $data;

 <?php var_dump($title)?>
  <?php var_dump($name)?
  <?php var_dump($picture)?>

我需要将这些值传递给视图，因此，我将上述代码修改为：

    class Upload extends CI_Controller
{
    function  __construct() {
        parent::__construct();
    }

public function input_values(){
    $data = array(
        'name' => $this->input->post('name',true),
        'picture' => $this->file_upload($_FILES['picture'])
    );
$this->load->view('documents', $data);    }
        function add(){
        $data = $this->input_values();
        if($this->input->post('userSubmit')) {
          $this->file_upload(($_FILES['picture']));
            if (!empty($_FILES['picture']['name'])) {
                $config['upload_path'] = 'uploads/docs/';
                $config['allowed_types'] = 'jpg|jpeg|png|gif|pdf|docx';
                $config['file_name'] = $_FILES['picture']['name'];
                $data['picture']=$this->file_upload($_FILES['picture']);
            }
        }

        return $this->db->insert('files', $data);
    }

    //logo image upload
    public function file_upload($file)
    {
        $this->my_upload->upload($file);
        if ($this->my_upload->uploaded == true) {
            $this->my_upload->file_new_name_body = 'file_' . uniqid();
            $this->my_upload->process('./uploads/docs/');
            $image_path = "uploads/docs/" . $this->my_upload->file_dst_name;
            return $image_path;
        } else {
            return null;
        }
    }

}

但我只能得到头衔的价值。名称和标题都出现以下错误：

Message: Undefined variable: name

我从视图中访问了变量，如下所示：

  $data = array(
    'title' => $this->input->post('title', true),
    'name' => $this->input->post('name',true),
    'picture' => $this->file_upload($_FILES['picture'])
);
return $data;

 <?php var_dump($title)?>
  <?php var_dump($name)?
  <?php var_dump($picture)?>

---

因此，这部分是获取post数据和加载视图（包含上载表单）的地方

然后这部分是处理上传表单的post请求：

function add() {
        $data = $this->input_values();
        if($this->input->post('userSubmit')) {
          $this->file_upload(($_FILES['picture']));
            if (!empty($_FILES['picture']['name'])) {
                $config['upload_path'] = 'uploads/docs/';
                $config['allowed_types'] = 'jpg|jpeg|png|gif|pdf|docx';
                $config['file_name'] = $_FILES['picture']['name'];
                $data['picture']=$this->file_upload($_FILES['picture']);
            }
        }

        return $this->db->insert('files', $data);
    }

这部分是你上传文件的地方

public function file_upload($file)
    {
        $this->my_upload->upload($file);
        if ($this->my_upload->uploaded == true) {
            $this->my_upload->file_new_name_body = 'file_' . uniqid();
            $this->my_upload->process('./uploads/docs/');
            $image_path = "uploads/docs/" . $this->my_upload->file_dst_name;
            return $image_path;
        } else {
            return null;
        }
    }

当您调用add（）函数时，它会调用input_values（）函数，然后加载视图，这样就不会执行下一行代码（cmiiw）

所以，也许你想改变一下：

public function index() {
   if ($this->input->post()) {
      // then handle the post data and files tobe upload here
      // save the post data to $data, so you will able to display them in view
   } else {
      // set the default data for the form
      // or just an empty array()
      $data = array();
   }

   // if the request was not a post, render view that contain form to upload file
   $this->load->view('nameOfTheView', $data);
}

---

你能显示你试图读取/访问这些值的代码吗？我刚刚对这些变量进行了var_转储请检查我的编辑请显示整个控制器。您没有给我们足够的信息，无法从中找到解决方案。您的问题标题具有误导性。请检查我编辑的代码