laravel 4分页收集 [英] laravel 4 paginate collection

问题描述

我无法使用laravel 4创建合适的分页系统.我具有以下返回集合的模型和函数:

I cant create a proper pagination system using laravel 4. I have the following models and function that return collections:

模型餐厅:

public function fooditem()
{
    return $this->hasMany('Fooditem','rest_id');
}
public function get_rest_foods($id){
    return Restaurant::find($id)->fooditem->toArray();
}

第二个函数以数组的形式返回某个餐厅的所有食物.我也在API调用中使用了它.

The second function returns all food items for a certain restaurant as an array. I also use this in an API call.

在我的控制器中，我有这个:

in my controller i have this:

 $food = $food->get_rest_foods($id);
 $paginator = Paginator::make($food, 10, 5);

我将分页器传递到视图，它显示了链接，但还显示了我来自食物阵列的所有商品.

I pass the paginator to the view and it shows the links ok but also shows all my item from the food array.

我尝试使用

public function get_rest_foods($id){
        return Restaurant::find($id)->fooditem->paginate(5);
    }

但是我得到一个错误:

FatalErrorException:错误:调用未定义的方法Illuminate \ Database \ Eloquent \ Collection :: paginate()

FatalErrorException: Error: Call to undefined method Illuminate\Database\Eloquent\Collection::paginate()

我在这个网站和许多其他网站上进行了搜索，但无法确定如何对集合进行分页.

I searched this and many other sites but cant understant how to paginate a collection.

感谢您的帮助

推荐答案

分页程序必须使用offset/limit语句获取通常从数据库查询中获得的项目. 因此，当您拥有一个包含所有项目的集合时，应该自己进行偏移/限制.

The paginator must get the items that it would normally get from a database query with an offset/limit statement. So when you have a collection with all items, you should do the offset/limit yourself.

    $food = $food->get_rest_foods($id);

    $page = 1;
    if( !empty(Input::get['page']) ) {
        $page = Input::get['page'];
    }

    $perPage = 15;
    $offset = (($page - 1) * $perPage);

    $food = Paginator::make($food->slice($offset,$perPage, true)->all(), $food->count(), $perPage);

这篇关于laravel 4分页收集的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！