我有這個視圖來搜索姓名或電話號碼。laravel 5中的建築物搜索功能
<form method="POST" action="{{url('/telephone/search')}}">
{!! csrf_field() !!}
<input type="text" name="search" placeholder="Name or Telephone no.">
<input type="submit" class="btn-btn-default" value="search">
路由重定向到該控制器:
public function search(Request $request)
{
$name=$request->get('search');
$search=telephone::where('name','like',$name)
->orWhere('Name','like',%$name)
->orWhere('Name','like',$name%);
->paginate(5);
return view('telephone.searchview',['search'=>$search]);
}
然後應該表明這樣的觀點:
<thead style='background-color:silver'><tr><td>S.N.</td><td>Name</td><td>Telephone No.</td><td>Mobile No.</td><td>Options</td></tr></thead>
@if($search==NULL)
No Results!
@else
@foreach ($search as $li)
<tr><td></td><td>{{$li->Name}}</td><td>{{$li->telephone}}</td><td>{{$li->mobile}}</td>
<td>here</td></tr></table>
@endforeach
{!! $search->render() !!}
@endif
得到了錯誤的orwhere語法,也沒有結果沒有顯示。
仍然問題如果沒有結果匹配.'@if($ search == NULL)沒有結果! @else' – Steve