我试图在“书籍”中打印出所有书籍名称及其相关类别。表和' book_categories'表。
但是我收到了这个错误:
尝试获取非对象的属性(查看:books.blade.php)
我的两个表格模型的名称:
- book.php中
- BookCategory.php
我的书'表结构:
id(int),book_name(string),book_category_id(int)
我的书#categories'表结构:
id(int),book_category(string)
控制器:
public function getBooks(){
$books = Book::all();
return View.make('book.books')->with('books', $books);
}
books.blade.php
@foreach ($books as $book)
<p>{{ $book->book_name }} </p>
<p>{{ BookCategory::find($book->id)->book_category }}</p>
@endforeach
只有这行代码才能正确打印出JSON信息:BookCategory::find($book-id)
但是当我使用
时- &gt;首先()
或
- &GT; book_category
要访问其内容,它无效。
答案 0 :(得分:1)
class Books extends \Model {
$table = 'books';
public function categories()
{
return $this->belongsTo('BookCategory');
}
}
class BookCategory extends \Model {
$table = 'book_categories';
public function books()
{
return $this->hasMany('Books', 'book_category_id');
}
}
public function getBooks()
{
// Always eagerload your relation to prevent n+1 issue
$books = Books::with('categories')->all();
return View::make('book.books')->with('books', $books);
}
@foreach($books as $book)
{{ $book->book_name}}
{{ $book->categories->book_category }}
@endforeach
进一步阅读:
http://laravel.com/docs/4.2/eloquent#querying-relations http://daylerees.com/codebright/eloquent-relationships