Question

我发誓我总是对CodeIgniter如何将控制器变量传递给他们的视图感到困惑，而我在其中执行的每个项目都会遇到同样的愚蠢问题。

这是我的模特：

public function get_post() {
   $sql = "SELECT * FROM posts WHERE active='y' AND head=1 LIMIT 1";

   $query = $this->db->query($sql);
   return $query->row();
}

这是我的控制器：

public function get_head() {
  $this->load->model('home_model');
  $data['head'] = $this->home_model->get_post();
  $this->load->view('home_view', $data);
}

当我在我的视图上执行var_dump（$ head）时，我得到：

object(stdClass)#18 (16) { ["id"]=> string(1) "1" ["member_id"]=> string(1) "1" ["title"]=> string(131) "This is a test title" ["slug"]=> string(0) "this-is-a-test-title" ["body"]=> string(0) "" ["tag"]=> string(7) "singles" ["orig_photo_name"]=> string(142) "sample.jpg" ["photo"]=> string(24) "14004803271605326007.jpg" ["comments"]=> string(1) "y" ["post_date"]=> string(19) "2014-06-17 02:18:47" ["head"]=> string(1) "1" ["views"]=> string(2) "50" ["notes"]=> string(0) "" ["nsfw"]=> string(1) "y" ["active"]=> string(1) "y" }

最后，当试图访问类似标题的内容时：

<?php echo $title; ?>

A PHP Error was encountered

Severity: Notice

Message: Undefined variable: title

Filename: views/home_view.php

Line Number: 8

除了切换到CakePHP之外的任何想法？

Answer 1

$query->row()将返回对象，因此请尝试访问视图

echo $head->id;
echo $head->title;

等等

Answer 2

First You change the model query like this 
public function get_post() {
   $sql = "SELECT * FROM posts WHERE active='y' AND head=1 LIMIT 1";
   $query = $this->db->query($sql);
   return $query->row_array(); //this is return value in array
}

Then call function in Controller

public function get_head() {
  $this->load->model('home_model');
  $data['head'] = $this->home_model->get_post();
  $this->load->view('home_view', $data);
}

Then get value in view 
<?php echo $head['id']; ?>

无法在CodeIgniter视图上访问控制器变量

2 个答案: