Laravel eloquent - 一对多关系（试图获取非对象的属性）
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Laravel eloquent - One to many relationships (Trying to get property of non-object)
提问by Ito Fachruzzaman
I have "posts" table that has many-to-one relationship with "categories" table. The goal is to show all of posts and their categories.
我有与“类别”表具有多对一关系的“帖子”表。目标是显示所有帖子及其类别。
Tables:
Posts: id, content, category_id, etc
Categories: id,name
表格：
帖子：id、内容、category_id 等
类别：id、名称
Here's my code
这是我的代码
Models:
楷模：
class Posts extends Eloquent
{
public static $table = 'posts';
public function categories()
{
return $this->belongs_to('Categories');
}
}
class Categories extends Eloquent
{
public static $table = 'categories';
public function posts()
{
return $this->has_many('posts');
}
}
My controller
我的控制器
public function get_posts()
{
$posts = Posts::with('categories')->all();
return View::make('admin.posts')
->with('title', 'Posts')
->with('posts', $posts);
}
My view
我的看法
@foreach($posts as $post)
<tr>
<td>{{ $post->title }}</td>
<td>{{ $post->categories->name }}</td>
<td><small> {{$post->updated_at}} </small></td>
<td>
<button>
{{HTML::link_to_route('edit_post','Edit',array($post->id))}}
</button>
{{Form::open('admin/delete','Delete')}}
{{Form::hidden('id', $post->id)}}
<input type="submit" name="edit_post" value="Delete"/>
{{Form::close()}}
</td>
</tr>
@endforeach
ERROR:
错误：
Error rendering view: [admin.posts]
Trying to get property of non-object
I am a newbie, please help me solve this issues
我是新手，请帮我解决这个问题
回答by Bonci Marco
{{ $post->categories->name }} before test is categories exists
{{ $post->categories->name }} 在测试之前是类别存在
Example:
例子：
@if( ! empty($post->categories))
<td>{{ $post->categories->name }}</td>
@else
@end if
Aesis.
艾西斯。
回答by Can Geli?
Just use ::all()instead of ::with(..)->all()
只需使用::all()代替::with(..)->all()
回答by adib.mosharrof
categories is an array as you are using a has_many relationship. There are many categories, hence an array is returned, so to access it you have to index it like an array
category 是一个数组，因为您正在使用 has_many 关系。有很多类别，因此返回一个数组，因此要访问它，您必须像数组一样索引它
the correct solution would be
正确的解决方案是
$post->categories[0]->name
$post->categories[0]->name
回答by Xiao Liu
Are you using Laravel 4? First the syntax for declaring relationship is hasMany and belongsTo, in camel case. Check it out in Laravel documentation
你在使用 Laravel 4 吗？首先声明关系的语法是hasMany 和belongsTo，在驼峰情况下。在Laravel 文档中查看
In view, check if categories are empty collection, ie., whether post has its category:
在视图中，检查类别是否为空集合，即帖子是否有其类别：
@if($post->categories->count())
<td>{{ $post->categories->name }}</td>
...
@endif
By the way, I would use singular form as model class name, like Post and Category instead of plural forms. And in Post class I would define inverse one-to-many relationship with singular form, to show there's only one entry in category table for this given Post.
顺便说一句，我会使用单数形式作为模型类名称，例如 Post 和 Category 而不是复数形式。在 Post 类中，我将定义具有单数形式的反向一对多关系，以显示此给定 Post 的类别表中只有一个条目。
public function category()
{
return $this->belongsTo('Category');
}
