我需要在 Laravel 中制作搜索表单
这是代码welcome.blade.php:
<form action="/search" method="POST" role="search">
{{ csrf_field() }}
<div class="input-group">
<input type="text" class="form-control" name="q"
placeholder="Search job"> <span class="input-group-btn">
<button type="submit" class="btn btn-default">
<span class="glyphicon glyphicon-search"></span>
</button>
</span>
</div>
</form>
<div class="container">
@if(isset($details))
<div class="row">
@foreach( $details as $job)
<h5>{{$job->company_name}}</h5>
<h3>{{$job->job_name}}</h3>
@endforeach
</div>
@endif
</div>
这是web.php中的路由:
Route::any('/search', function () {
$q = Input::get('q');
$job = Job::where('job_name', 'LIKE', '%' . $q . '%')
->orWhere('company_name', 'LIKE', '%' . $q . '%')
->get();
if (count($job) > 0)
return view('welcome')
->withDetails($job)
->withQuery($q);
else
return view('welcome')
->withMessage('No Details found. Try to search again !');
});
我收到错误:
调用未定义的方法 Symfony\Component\Console\Input\Input::get()
我该如何修复它?
我还想知道,如何才能先显示所有职位,然后再搜索匹配的职位?
临摹微笑
桃花长相依