Question

所以这是我的控制器：

        class PostsController extends BaseController
    {
        public function postSearch()
        {
        $q = Input::get('username');

        $posts = DB::table('users')->whereRaw(
            "MATCH(username) AGAINST(? IN BOOLEAN MODE)",
            array($q)
        )->get();

        return View::make('posts.index', compact('posts'));

    }


}

和我的路线：

Route::get('posts/index', function()
{
    return View::make('posts/index');
});


Route::post(
    ''posts/index',
    array(
        'as' => 'posts.index',
        'uses' => 'PostsController@postSearch'
    )


);

和我的HTML：

<div class="search">
    {{ Form::model(null, array('route' => array('posts.index'))) }}
    {{ Form::text('username', null, array( 'placeholder' => 'Search query...' )) }}
    {{ Form::submit('Search') }}
    {{ Form::close() }}


</div>

问题是它给我错误：

MethodNotAllowedHttpException

所以我不知道用户能够在索引页面中搜索并在表格的diffiere页面中给出结果如何显示结果以及这个错误是什么？

Answer 1

你做的一切都很好，但是在路径文件中你的类被命名为PostsController，但实际的控制器类名是postController这就是为什么它没有找到PostsController。

<强>路线：

Route::get('posts/index', function()
{
    return View::make('posts/index');
});

Route::post('posts/index', array('as' => 'posts.index','uses' => 'PostsController@postSearch'));

<强>控制器

class PostsController extends BaseController
{
    public function postSearch()
    {
        $q = Input::get('username');

        $posts = DB::table('users')->whereRaw(
            "MATCH(username) AGAINST(? IN BOOLEAN MODE)",
            array($q)
        )->get();

        return View::make('posts.index', compact('posts'));

    }
}

查看

<div class="search"> {{ Form::open(array('route' => 'post.index', 'method' => 'POST', 'role' => 'search')) }} {{ Form::text('username',null, array( 'placeholder' => 'Search query...' )) }} {{ Form::submit('Search') }} {{ Form::close() }} </div> @if(isset($posts)) @foreach($posts as $post) {{$post->id}} @endforeach @endif

laravel中的搜索引擎无法正常工作+功能

1 个答案: