從搜索表單中搜索一個參數，可以帶多個參數Laravel

Question

我創建了一個搜索表單，可以有3個參數來搜索數據庫並返回結果。表單中的所有參數或表單字段都是下拉菜單。

表單代碼：

{!!Form::open(['route' => 'search','class' => 'form-inline pull-left', 'id' => 'search-form', 'method' => 'GET', 'role' => 'search'])!!} 
    <select class=" form-control" name="country"> 
     <option> Select Country</option> 
     @foreach($countries as $country) 
     <option value="{{$country->name}}">{{ucfirst($country->name)}}</option> 
     @endforeach 
    </select> 
    <select class=" form-control" name="category"> 
     <option value="">Select Activity</option> 
     @foreach($categories as $category) 
     <option value="{{$category->name}}">{{ucfirst($category->name)}}</option> 
     @endforeach 
    </select> 
    <select class=" form-control" name="days"> 
     <option value="">Select Duration (Days)</option> 
     <option value="1|5">1-5</option> 
     <option value="10|15">10-15</option> 
     <option value="20|30">20-30</option> 
     <option value="30|60">30-Above</option> 
    </select> 
{{ Form::submit('Search', ['class' => 'btn btn-primary btn-lg','id' => 'search'])}} 

    {!! Form::close() !!} 

我在控制器的方法只有一個參數來搜索：

public function search(Request $request) 
{ 
    $days = $request->days; 
    $days_explode = explode('|', $days); 

    if (isset($request->country) && !isset($request->category) && !isset($request->days)) { 

     $query= Tour::whereHas('country', function($r) use($request) { 
      $r->where('name','=', $request->country); 
     }) 
     ->paginate(8); 

     return view('public.tour.list')->withResults($query); 
    } 
    else{ 
    $query= Tour::whereHas('country', function($r) use($request) { 
      $r->where('countries.name','=', $request->country); 
     }) 
     ->whereHas('category', function($s) use($request) { 
      $s->where('categories.name','=', $request->category); 
     }) 
     ->whereHas('country', function($r) use($request) { 
      $r->where('countries.name','=', $request->country); 
     }) 
     ->whereBetween('days', [$days_explode[0], $days_explode[1]]) 
     ->paginate(8); 
    return view('public.tour.list')->withResults($query); 
    } 

我試圖讓我的形式靈活。它應該返回1,2或3參數是否傳遞給它的結果。使用上面的代碼，如果所有3個參數都傳遞給它，表單將返回結果。如果只有一個參數傳遞給它，則會給出錯誤Undefined offset: 1。

Answer 1

您可以先檢查每個參數，然後決定是否將其添加到查詢中。做這樣的事情：

$query = Tour::where(function($q) use($request) { 
    if (!empty($request->country)) { 
     $q->whereHas('country', function($r) use($request) { 
      $r->where('countries.name', $request->country); 
     }); 
    } 

    if (!empty($request->category)) { 
     $q->whereHas('category', function($r) use($request) { 
      $r->where('categories.name', $request->category); 
     }); 
    } 
}) 
->whereBetween('days', [$days_explode[0], $days_explode[1]]) 
->paginate(8); 

或者你可以這樣做：

$query = Tour::query(); 

if (!empty($request->country)) { 
    $query = $query->whereHas('country', function($r) use($request) { 
     $r->where('countries.name', $request->country); 
    }); 
} 

if (!empty($request->category)) { 
    $query = $query->whereHas('category', function($s) use($request) { 
     $s->where('categories.name', $request->category); 
    }); 
} 

$query = $query->whereBetween('days', [$days_explode[0], $days_explode[1]]) 
       ->paginate(8);