包含多个条件的Laravel搜索查询 - Laravel search query with multiple conditions

PHP/Laravel新手，请耐心等待。

我有一个网页，是基于3个标准搜索狗，品种，性别和半径。

下面是相关代码:

搜索页面

<div class="col-md-12 zero-pad-left zero-pad-right"> 
{{ Form::open(array('action' => array('DogsController@index'), 'class'=>'form width88', 'role'=>'search', 'method' => 'GET')) }} 
<div id="prefetch">
  {{ Form::text('search-breed', null, array('class' => 'typeahead form-group form-control', 'placeholder' => 'Search by breed here...')) }}
  {{ Form::text('sex', null, array('class' => 'form-group form-control', 'placeholder' => 'Search by sex here...')) }}
  {{ Form::text('miles', null, array('class' => 'form-group form-control', 'placeholder' => 'Search by distance here...')) }}


</div>
{{ Form::submit('Search', array('class' => 'btn btn-default search-bar-btn')) }}
{{ Form::close() }}

ControllerPage

class DogsController extends 'BaseController {
public function __construct()
{
    // call base controller constructor
    parent::__construct();
    // run auth filter before all methods on this controller except index and show
    $this->beforeFilter('auth', array('except' => array('show')));
}
/**
 * Display a listing of the resource.
 *
 * @return Response
 */
public function index()
{
    if (Input::has('search')) {
        $queryString = Input::get('search');
        $dogs = Dog::where('name', 'LIKE', "%$queryString%")->orderBy('name')->paginate(5);
    }
    elseif (Input::has('search-breed')) 
    {
        $dogs = Dog::whereHas('breed', function($q)     
        {
            $queryString = Input::get('search-breed');
            $q->where('name', 'LIKE', "%$queryString%");
        })->orderBy('name')->paginate(5);
    } //end elseif
    else {
        $dogs = Dog::orderBy('name')->paginate(5);
        } //end else
    return View::make('dogs.index')->with(array('dogs' => $dogs));
} //end function index()

当我输入搜索"poodle, male, within 20 miles"时，url显示如下:

http://ruff-love.dev/dogs?search-breed=poodle&sex=M&miles=20

当搜索仅繁育时，搜索当前工作正常。

我似乎也弄不明白添加SEX和RADIUS标准的语法。它应该允许这些条件为空，并且仍然执行查询。

如有任何建议，不胜感激

您可以使用查询范围http://laravel.com/docs/eloquent#query-scopes使其在您的控制器中变得冗长和更容易(或将来您将在任何地方这样做)，然后根据您的需要链接它们:

// Dog model
public function scopeSearchBreed($query, $breed)
{
  $query->whereHas('breed', function ($q) use ($breed) {
    $q->where('name', 'like', "%{$breed}%");
  });
}
public function scopeWithinRadius($query, $radius)
{
  $query->where(...); // do math here
}

那么你所需要的就是:

public function index()
{
  $q = Dog::query();
  if (Input::has('search'))
  {
     // simple where here or another scope, whatever you like
     $q->where('name','like',Input::get('search'));
  }
  if (Input::has('search-breed'))
  {
     $q->searchBreed(Input::get('search-breed'));
  }
  if (Input::has('sex'))
  {
     $q->where('sex', Input::get('sex'));
  }
  if (Input::has('radius'))
  {
     $q->withinRadius(Input::get('radius'));
  }
  $dogs = $q->orderBy(..)->paginate(5);
  // ...

这是一个可能的解决方案，我想可能还有其他的。使用query()函数创建一个空查询构建器，并向其添加非空子句，然后在末尾调用paginate()函数。

$builder = Dogs::query();
if (Input::has('search')) {
    $queryString = Input::get('search');
    $builder->where('name', 'LIKE', "%$queryString%");
}
// ... more clauses from the querystring
$dogs = $builder->orderBy('name')->paginate(5);

$builder = Dogs::query();
$term = Request::all();
if(!empty($term['breed'])){
    $builder->where('breed','=',$term['breed']);
}
if(!empty($term['sex'])){
    $builder->where('sex','=',$term['sex']);
}
if(!empty($term['radius'])){
    $builder->where('radius','=',$term['radius']);
}
$result = $builder->orderBy('id')->get();

$area = Area::query();
if (Input::has('search')) {
    $queryString = Input::get('search');
    $area->where('name', 'LIKE', "%" . $queryString . "%");
}
$result = $area->orderBy('name')->paginate(5);