如何在laravel 5中发布ajax数据?

时间:2018-01-06 09:57:27

标签: jquery ajax laravel laravel-5

这是我的路线档案

interface LayoutInterface {
    getLayout();
}

class LayoutEnglish implements LayoutInterface {
    getLayout() {
        return {'Q', 'W', 'E', 'R'...};
    }
}

class LayoutNumeric implements LayoutInterface {
    getLayout() {
        return {'1', '2', '3', ...};
    }
}

class LayoutComponent {
   @Input
   type: 'english';

   getLayout(type) { 
     // what about dynamic import here?
     return new {Layout+type};
   }
}


keypad.component.html

<app-keypad type="english"></app-keypad>

这是我的jquery ajax代码

Route::post('/ajaxdemo', 'DemoController@saveAjax');

当我提交表单时,ajax请求转到控制器方法

$('#record_form').on('submit', function (e) {

    e.preventDefault();

    var select_form = $('#select_form').serialize();

    $.ajax({
        type: "POST",
        url: '/ajaxdemo',
        data: {
            mydata: select_form,
        },
        dataType: 'json',
        success: function (data) {

            console.log(data);

        }
    });
});

我从public function saveAjax(Request $request) { $data = $request->all(); dd($data); } 调用ajax函数,页面只是刷新并在/demo/create中显示token。即使我正在做dd($ data)

,它也不会显示数据

3 个答案:

答案 0 :(得分:0)

请试试这个:

 <input type="text" id="result">//treat this as location 2
 <div id="map"  style="width:100%;height:200px;background:yellow;"    frameborder="0" style="border:0"></div>

 <script>
 function myMap() {
var mapOptions = {
center: new google.maps.LatLng(15.835015,74.517537),
zoom: 25,
mapTypeId: google.maps.MapTypeId.HYBRID
 }
  var map = new google.maps.Map(document.getElementById("map"), mapOptions);
     var marker = new google.maps.Marker({
      position: GIT,
      map: map
    });

 }
</script>


<script src="https://maps.googleapis.com/maps/api/js?key=&callback=myMap"></script>
</input>


</form>

答案 1 :(得分:0)

$.ajax({
      url:$("#select_form").attr('action'),
      dataType:'json',
      type:'POST',
      data:$("#select_form").serialize(),
      success: function(data) {}        
     });

答案 2 :(得分:0)

如果你想通过ajax laravel显示数据。你可以通过打开inspect-&gt;网络检查数据响应或dd($ data)并再次运行ajax。你可以看到任何数据或错误

也许您的请求错误500可由https://laravel.com/docs/5.5/csrf

修复