svil4ok/laravel-query-filters

1.0.1 2017-11-14 13:41 UTC

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想象一下,如果我们需要根据某些标准(如名称、电子邮件、位置、公司等)过滤存储在数据库中的所有用户,将会怎样。

/users?country=Bulgaria&city=Sofia&name=Svilen

例如,为了过滤所有这些参数,我们可能需要执行类似以下操作:

<?php

namespace App\Http\Controllers;

use Illuminate\Http\Request;
use App\Http\Requests;
use App\User;

class UserController extends Controller
{
    public function index(Request $request)
    {
        $users = (new User)->newQuery();

        if ($request->has('country')) {
            $users->where('country', '=', $request->get('country'));
        }

        if ($request->has('city')) {
            $users->where('city', '=', $request->get('city'));
        }
        
        if ($request->has('name')) {
            $users->where('name', 'LIKE', '%' . $request->get('name') . '%');
        }

        // ...
        // other filters
        // ...

        return $users->get();
    }
}

使用这个包,您可以轻松地根据请求的查询字符串创建过滤器,并将控制器重构为以下形式:

<?php

namespace App\Http\Controllers;

use Illuminate\Http\Request;
use App\Http\Requests;
use App\User;

class UserController extends Controller
{
    public function index(UserFilters $filters)
    {
        return User::filter($filters)->get();
    }
}

用法

  • Filterable特质添加到您的模型中,以允许使用Model::filter()
<?php

namespace App;

use Illuminate\Database\Eloquent\Model;
use SGP\QueryFilters\Filterable;

class User extends Model
{
    use Filterable;
}
  • 使用以下模板生成您的模型过滤器:
<?php

namespace App;

use SGP\QueryFilters\QueryFilters;

class UserFilters extends QueryFilters
{
    public function filterByOption($value)
    {
        return $this->builder->where('column', 'operator', $value);
    }
}
  • 使用您的过滤器
<?php

namespace App\Http\Controllers;

use App\User;
use App\UserFilters;

class MyController extends Controller
{
    public function index(UserFilters $filters)
    {
        return User::filter($filters)->get();
    }
}