【发布时间】:2016-12-11 11:27:59
【问题描述】:
我想对 Union 查询结果进行分页。我写了这个:
use Illuminate\Support\Facades\Input;
use Illuminate\Pagination\Paginator;
$page = Input::get('page', 1);
$paginate = 10;
$members = DB::table("members")
->select("id", "first_name", "last_name", "email", "created_at")
->where("site_id", $id);
$users = DB::table("users")
->select("id", "first_name", "last_name", "email", "created_at")
->where("site_id", $id)
->union($members)
->get()->toArray()
$slice = array_slice($users, $paginate * ($page - 1), $paginate);
$data = new Paginator($slice, $paginate);
return View::make('main.pages.search',compact('data','searchTerm'));
现在我想访问search刀片模板中的结果。假设我写了这个:
<div class="container">
@foreach ($data as $user)
{{ $user->first_name }}
@endforeach
</div>
{{ $data->links() }}
但是我收到了这个错误:
Trying to get property of non-object (View: D:\wamp\www\aids\resources\views\main\pages\search.blade.php
无法识别$user,它是name 属性
什么是问题?
更新:
而上面的return $data 则返回这个:
{
"per_page": 2,
"current_page": 1,
"next_page_url": null,
"prev_page_url": null,
"from": 1,
"to": 2,
"data": [
{
"id": 6,
"first_name": "ali",
"last_name": "hassani",
"created_at": "2012-04-16 22:11:46",
"email": "ali@gmail.com",
},
{
"id": 7,
"first_name": "hossein",
"last_name": "rezaei",
"created_at": "2012-04-16 22:11:46",
"email": "reza@gmail.com",
},
]
}
我尝试$data->data 访问结果,但出现同样的错误。
更新 2:
我发现当我像这样添加{{dd($user)}} 时:
<div class="container">
@foreach ($data as $user)
{{dd($user)}}
{{ $user->first_name }}
@endforeach
</div>
返回 array 包含 Result 而不是 object。
【问题讨论】:
-
您选择了 first_name 和 last_name 但您试图显示
$user->name? -
我尝试过,但出现了同样的问题。
-
var_dump($data)并检查它是否没有添加新的深度级别。
标签: php laravel pagination laravel-5.3