如何在laravel中对相关模型进行分页

Posted 2023-03-14

【中文标题】如何在laravel中对相关模型进行分页

【英文标题】：How to paginate a related model in laravel

【发布时间】：2015-08-10 14:39:35

【问题描述】：

我有两个模型，即 RoomCategory 和 Room。我想通过foreach循环对每个房间类别相关的所有房间进行分页，这是我的MVC。

型号

class RoomCategory extends \Eloquent 
    public function Room()
        return $this->hasMany('Room','category_id','id','RoomCategory');
    


class Room extends \Eloquent 
    public function RoomCategory()
        return $this->belongsTo('RoomCategory','id','category_id','Room');
    

控制器

public function index()

    $room_category = RoomCategory::get()
        ->with('Room')->paginate(1);

    return View::make('pages.roomlist', compact('room_category'));

查看

@foreach($room_category as $category)
    $category->name
    @foreach($category->room as $rooms)

         $rooms->name

         $category->room->links()
    @endforeach
@endforeach

我得到了这个错误

调用未定义的方法 Illuminate\Database\Eloquent\Collection::with()

有人可以帮助我如何在房间类别的 foreach 循环下对这些房间进行分页。谢谢

【参考方案1】：

如果您使用分页，则不需要 get() 方法。替换这个

$room_category = RoomCategory::get()
    ->with('Room')->paginate(1);

有了这个

$room_category = RoomCategory::with('Room')->paginate(1);

编辑

最好不要在索引页面中使用分页链接来表示房间，只需对房间类别进行分页即可。

$room_category = RoomCategory::with([
                'Room' => function($query) 
                    $query->latest()->limit(5);
                
                ])->paginate(3);

在您的索引视图上对类别进行分页，并为每个类别页面建立一个链接并将您的房间分页

index.blade.php

@foreach($room_category as $room_cat)
    @foreach($room_cat->Room as $room)
        <div class="room-section"></div>
    @endforeach
    <a href=" url('category', $room_cat->id) ">View more of this category</a>
@endforeach

 $room_category->links() 

你的控制器

public function showCategory($id)

    $room_category = RoomCategory::with([
                'Room' => function($query) 
                    $query->latest()->paginate(8);
                
                ])->find($id);

    return View::make('category.show', compact('$room_category'));

和category/show.blade.php你可以对房间进行分页

@foreach($room_category->Room as $room)
    <div class="room-section"></div>
@endforeach

 $room_category->Room()->links() 

【讨论】：

在我执行此代码时再次出错Call to undefined method Illuminate\Database\Eloquent\Collection::links()

它分页 $room_category 而不是 $category->room

【参考方案2】：

要实现您想要的，您必须获取所有房间类别，遍历它们，然后遍历每个房间类别房间。

这是示例代码：

public function index()

    $room_category = RoomCategory::with('Room')->paginate(1);

    return View::make('pages.roomlist', compact('room_category'));

你的看法：

@foreach($room_category as $category)
    $category->name
    @foreach($category->Room as $rooms)
        $rooms->name
    @endforeach
@endforeach