【发布时间】:2014-05-08 21:22:29
【问题描述】:
所以我想要在我的视图中使用分页,但我总是收到错误Call to a member function paginate() on a non-object,这是我的代码:
首先我的行动:
public function index()
{
$logs = DB::table('services')
->join('logs', function($join)
{
$join->on('services.id', '=', 'logs.service_id');
})
->get()->paginate(10); //here i got the error
return View::make('logs.index',array('logs'=> $logs,'title'=>'Service Logs'));
}
那么我的看法:
<table class="table table-striped table-bordered">
<thead>
<tr>
<td>Domain</td>
<td>Port</td>
<td>Checktime</td>
<td>Status</td>
<td>Responce</td>
</tr>
</thead>
<div class="container">
@foreach($logs as $key => $value)
<tr>
<td>{{ $value->domain }}</td>
<td>{{ $value->service_port }}</td>
<td>{{ $value->checktime }}</td>
<td class="text-center">
@if( $value->status == 'up' ) <img src="../img/up3.png" />
@elseif( $value->status == 'down' ) <img src="../img/down3.png" />
@else <img width="30" height="30" src="../img/warning_icon.png" />
@endif
</td>
<td>{{ $value->response_time }}</td>
</tr>
@endforeach
</div>
</table>
{{$logs->links();}}
所以如果有人有任何想法,我会非常感激
【问题讨论】:
-
你得到什么错误?
-
在非对象上调用成员函数 paginate()
标签: php twitter-bootstrap laravel pagination laravel-4