在 Laravel 中搜索部分匹配项

Question

我在 Laravel 中编写了一个搜索功能，但我只能搜索完整的查询，这不是一个理想的解决方案。我的意思是，如果我通过像“lorem%20ipsum”这样的查询，我只会得到“lorem ipsum”的结果，而不会像“lorem”或“ipsum”那样得到部分匹配。谁能给我一些关于如何最好地完成这项任务的想法？

这是我目前完整的搜索代码：

Routes.php

Route::get('{lang}/search/{query}', 'HomeController@searchPages');

HomeController.php

public function searchPages($lang, $query) {

    $searchResults = Search::acme($query, $lang);

    return View::make('search.search')
    ->with('searchResults', $searchResults);
}

模型/Page.php

class Page extends Eloquent {

    public function scopeSearch($query, $search) 
    {
        return $query->where(function($query) use ($search) 
        {
            $query->where('title','LIKE', "%$search%")
                  ->orWhere('body', 'LIKE', "%$search%");
        });
    }
} 

Acme/Facades/Search.php

namespace Acme\Facades;

use Illuminate\Support\Facades\Facade;

class Search extends Facade {

    protected static function getFacadeAccessor() 
    {
        return 'search';
    }
}

Acme/Search/SearchServiceProvider.php

namespace Acme\Search;

use Illuminate\Support\ServiceProvider;

class SearchServiceProvider extends ServiceProvider {

    public function register()
    {
        $this->app->bind('search', 'Acme\Search\Search');
    }
}

Acme/Search/Search.php

namespace Acme\Search;

use Illuminate\Support\Collection;
use Page;

class Search {

    public function pages($search) 
    {
        return Page::search($search)->get();
    }

    public function acme($query, $lang) 
    {
        return new Collection(Page::join('langs', 'langs.id', '=', 'pages.lang_parent_id')
        ->where('title', 'LIKE', '%'.$query.'%')
        ->orWhere('body', 'LIKE', '%'.$query.'%')
        ->where('code', '=', $lang)
        ->get()
        ->toArray());
    }
}

Answer 1

你需要遍历你的搜索文本并建立你的口才，例如：

$db_query = Page::join('langs', 'langs.id', '=', 'pages.lang_parent_id');
$my_search_text_arr = explode(' ',$query);

$first = true
for_each($my_search_text_arr as $my_search_text){
   if($first){
      $db_query = $db_query->where('title', 'LIKE', '%'.$my_search_text.'%');
      $first = false;
   }else{
      $db_query = $db_query->orWhere('title', 'LIKE', '%'.$my_search_text.'%');
   }
}

 return new Collection($db_query
        ->where('code', '=', $lang)
        ->get()
        ->toArray());