我有两个表User &文章表之间的关系
模型:
class Article extends Eloquent {
public static $table = 'article';
public function User()
{
return $this->has_one('user', 'id');
}
和
class User extends Eloquent {
public static $table = 'user';
public function Article()
{
return $this->belongs_to('article', 'id_user');
}
我想从用户直接在文章视图上获得名称值，但不与错误工作试图获取非对象的属性
我的控制器:
public function action_index()
{
$Article = Article::order_by('id')->paginate(10);
return View::make('article.index')->with('$articles', $Article);
}
我的观点:
@foreach ($articles->results as $Arti)
<tr>
<td>{{$Arti->id}}</td>
<td>{{$Arti->tag}}</td>
<td>{{$Arti->user->name }}</td> <------ ERROR
<td>{{$Arti->content}}</td>
<td>{{$Arti->date}}</td>
<td>
看一下下面，有一些东西和你的不一样…
- Article belongs_to User (not has_one)
- User has_many Article (not belongs_to)
- 您的关系应该以小写命名，复数has_many(即
articles或user) - 您的关系主题应该是类名(即
Article或User) - 外键应该是名称relationship_id，即
user_id - 将
::with()添加到您的查询中 - 当你分页时，你需要在视图中访问
->results
class Article extends Eloquent {
// 3: lowercase 'user'
public function user()
{
// 1: Article belongs to User
// 4: class name 'User'
// 5: Foreign key on article table is user_id
return $this->belongs_to('User');
}
}
// models/user.php
class User extends Eloquent {
// 3: lowercase plural 'articles'
public function articles()
{
// 2: User has many Articles
// 4: class name 'Article'
return $this->has_many('Article');
}
}
// controllers/articles.php
class Article_Controller extends Base_Controller {
public $restful = true;
public function get_index()
{
// 6: Eager load the user relationship, ::with('user')
$articles = Article::with('user')->order_by('id')->paginate(10);
return View::make('articles.index', compact('articles'));
}
}
// views/articles/index.blade.php
// 7: access $articles->results from the paginator
@foreach ($articles->results as $article)
<h1>{{ $article->title }}</h1>
<p>By {{ $article->user->name }}</p>
@endforeach
{{ $articles->links() }}