Laravel 发布 + ajax

Question

我在使用 laravel 时遇到问题，我正在运行一个 SPA，我需要创建一个按类别搜索和过滤的表单。

我正在通过 POST 执行此操作，但我需要使用过滤器和 ajax 再次运行查询，以便页面不会重新加载。

但是我该怎么做呢？我只需要运行该查询并重新加载 div，ajax 部分还可以，但我不知道如何在不运行每个查询以“填充”页面内容的情况下仅运行该查询。

对不起，如果不够清楚，我的英语不太好。

感谢您的建议！

控制器：

public function index()
{
    $data['servicos']                   = DB::select('SELECT * FROM servicos WHERE ativo = 1 ORDER BY ORDEM ASC');
    $data['servicos_processo']          = DB::select('SELECT * FROM servicos_processo WHERE ativo = 1 ORDER BY ID ASC');
    $data['servicos_fun']               = DB::select('SELECT * FROM servicos_fun WHERE ativo = 1 ORDER BY ID ASC');
    $data['curriculum_historico']       = DB::select('SELECT * FROM curriculum_historico WHERE ativo = 1 ORDER BY ID DESC');
    $data['curriculum_educacao']        = DB::select('SELECT * FROM curriculum_educacao WHERE ativo = 1 ORDER BY ANO ASC');
    $data['curriculum_coding_skills']   = DB::select('SELECT * FROM curriculum_coding_skills WHERE ativo = 1 ORDER BY ID ASC');
    $data['curriculum_other_skills']    = DB::select('SELECT * FROM curriculum_other_skills WHERE ativo = 1 ORDER BY ID ASC');
    $data['curriculum_testimonials']    = DB::select('SELECT * FROM curriculum_testimonials WHERE ativo = 1 ORDER BY ID ASC');
    $data['certificados_last']          = DB::select('SELECT * FROM certificados WHERE ativo = 1 ORDER BY data_conclusao DESC LIMIT 9');
    return view('main', $data);
}

public function certificados()
{
    $nome = Request::input('filtro-certificados');
    $data['certificados_last']          = DB::select("SELECT * FROM certificados WHERE ativo = 1 AND nome LIKE '%{$nome}%' ORDER BY data_conclusao DESC LIMIT 9");
    return $data;
}

查看：

 <div class="latest-posts media-grid masonry" data-layout="masonry" data-item-width="340">

                @foreach($certificados_last as $certificado)
                <article class="hentry media-cell">
                    <div class="media-box">
                        <img src="images/certificados/{{$certificado->id}}.svg" style="padding:60px" alt="post-image">
                        <div class="mask"></div>
                        <a href="certificado/{{$certificado->id}}"></a>
                    </div>

                    <header class="media-cell-desc">
                            <span title="2013" class="date">
                                <span class="day">{{$certificado->dia_conclusao}}</span>{{$certificado->mes_conclusao}}</span>
                        <h3>
                            <a href="certificado/{{$certificado->id}}">{{$certificado->nome}}</a>
                        </h3>
                    </header>

                </article>
                @endforeach

            </div>

Answer 1

使用 axios、jQuery 或您选择的任何客户端发送发布请求。包括您的搜索和过滤参数。

在 Laravel 中，创建指向控制器方法的 post 路由。这个方法应该直接返回数据——不要返回视图。

例如：

public function ajaxSearch(Request $request)
{
    $data = User::where('name', $request->query);

    return $data;
}

然后，使用您的 JavaScript 替换或插入新数据。

• 如果您还没有，请查看 Vue。

注意：Laravel 会自动将响应转换为 JSON。

使用done() 方法，就像使用 jQuery 时一样：

$.post( '/myRouteHere', { name: 'John', time: '2pm' })
    .done(function() {

        // insert / update data

    }).fail(function() {

        alert( "error" );

    });