CakePHP-2.0如何从$ allposts = $ this-> paginate（'Post'）获取用户名;其中user_id是posts表的外键？ [英] CakePHP-2.0 How can i get the username from $allposts=$this->paginate('Post'); where user_id is foreign key of posts table?

问题描述

我使用CakeDC-Users插件。

I'm using CakeDC-Users plugin.

<?php
class Post extends AppModel { 
    public $useTable='posts';
    public $belongsTo = array('User');
    public $hasMany=array('Comment');
}

我不得不使用paginate：

I had to use paginate:

$ allposts = $ this-> paginate（'Post'）;

以这种方式获取user_id：

I can get the user_id in this way:

foreach ($allposts as $post) {
    debug($post['Post']['user_id']);

但我需要用户名而不是user_id。如何获取用户名？

But i need the username not the user_id. How can i get username?

推荐答案

CakPHP的containble功能默认隐藏所有关联模型： http://book.cakephp.org/2.0/en/core-libraries/behaviors/containable。 html

The containble feature of CakPHP hides all associated models by default: http://book.cakephp.org/2.0/en/core-libraries/behaviors/containable.html

如果您只想添加一个关联模型的一个字段，可以使用以下语法：

If you want to add only one field of an associated Model you can use this syntax:

$allposts = $this->Post->find('all', array('contain' => 'User.username'));

或者使用paginate在你的控制器类中使用this：（http://book.cakephp.org/ 1.3 / en / view / 1232 / Controller-Setup）

or with paginate use this in your controller class: (http://book.cakephp.org/1.3/en/view/1232/Controller-Setup)

var $paginate = array('contain' => 'User.username');

尝试以下方式访问它：

$post['User']['username'];

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