Question

我是新手，我已经陷入了让我在过去一天陷入困境的僵局

我有3个表帖子，类别和具有1-many关系的用户，其中一个类别可以有很多帖子。

类别表只有两个字段：id和cat_desc，id被设置为主键

在我的postscontroller中我有一个添加功能

public function add() {
    if ($this->request->is('post')) {

        $data = $this->Post->Cat->find('list', ['fields' => array('cat_desc')]);
        $this->set('cats', $data);

        $this->request->data['Post']['user_id'] = $this->Auth->user('id');

          if ($this->Post->save($this->request->data)) {
             $this->Session->setFlash(__('Your post has been saved.'));
             return $this->redirect(array('action' => 'index'));
           }
    }

    $this->set('title_for_layout',"'Add a Post");
    }

在我的视图中，我尝试调用下拉列表

      echo $this->form->create('Post', array('action'=>'add'));
      echo $this->form->input('title');

      echo $this->Form->input('category', array('options' => array('options'=> $cats))
      );

      echo $this->form->input('amount');
      echo $this->form->input('body');
      echo $this->form->end(__('Create a Post', true));

我得到一个未定义的变量猫。任何帮助将不胜感激，我感谢你们所有的帖子，因为它帮助我达到了这一点：）

Answer 1

首次导航到该页面时，请求类型将为＆＃34; get＆＃34;。代码的方式是，只有在请求类型为＆＃34; post＆＃34; （所以它只会在你提交表单时定义。）只需将设置$ cats的代码移到if语句if($this->request->is('post'))

之外。

Answer 2

//controller
    public function add() {
        if ($this->request->is('post')) {

            $this->request->data['Post']['user_id'] = $this->Auth->user('id');

              if ($this->Post->save($this->request->data)) {
                 $this->Session->setFlash(__('Your post has been saved.'));
                 return $this->redirect(array('action' => 'index'));
               }
        } else {

        $cats = $this->Post->Cat->find('list', ['fields' => array('cat_desc')]);
        $this->set('cats', $cats);
        $this->set('title_for_layout',"'Add a Post");
         }
        }

//视图

 echo $this->Form->input('cat_id', $cats, array(options here));

试图为下拉列表传递列表但找不到变量

2 个答案: