Question

有人可以在刀片模板中显示Laravel 5.7 post ajax示例以及完整的最小示例吗？我知道网络上有一些资源，但是我想念一个简洁，直接的最小示例。

Answer 1

您可以做这样的事情， web.php

  Route::post('/admin/order/{id}', 'OrderController@edit')->name('admin.order.edit');

blade.php

 $(document).on('click', '.delete-button', function (e) {
        e.preventDefault();
        var orderId = 1
        $.ajax({
            type: 'post',
            url: '/admin/order/' + orderId,
            data: {
                '_token': $('input[name=_token]').val(),
                 'data_one': 'dataone',
            },
            success: function () {     
                toastr.success('Order Has Been Deleted Successfully.');
            },
            error: function(XMLHttpRequest) {
                toastr.error('Something Went Wrong !');
            }
        });

    });

Answer 2

定义网络路由

Route::get('currencies/fiat/changeStatus','FiatCurrencyController@changeStatus')->name("currencies.fiat.chanageStatus");

在单击onclick =“ changeStatus（1,0）”时调用此函数

function changeStatus(id,status){

            var CSRF_TOKEN = $('meta[name="csrf-token"]').attr('content');
            $.ajax({
                /* the route pointing to the post function */
                url: '/currencies/fiat/changeStatus',
                type: 'GET',
                /* send the csrf-token and the input to the controller */
                data: {_token: CSRF_TOKEN,cid:id,status:status},
                dataType: 'JSON',
                /* remind that 'data' is the response of the AjaxController */
                success: function (data) {
                    console.log(data);



                }
            });
        }

仅此而已。

Answer 3

$(document).ready(function(){
/* In laravel you have to pass this CSRF in meta for ajax request  */
$.ajaxSetup({
  headers: {
      'X-CSRF-TOKEN': $('meta[name="csrf-token"]').attr('content')
  }
});

/* change in website layout base on click */
$('#user_view').on('click',function (e){
    e.preventDefault();
    $('.loading_screen').show();
    var val = $(this).data('id');
    $.ajax({
        url: base_path + 'user/changeview/'+val,
        type: "POST",
        success: function (data) {
            var obj = jQuery.parseJSON(data);
            if (obj.SUCC_FLAG == 0){
              window.location.href = site_url;}
          else{
              /* for console error message. */
              console.log(obj.MSG);}
          $('.loading_screen').hide(); 
       },
       error: function () {
          alert("server error");
         }
    });
});

}）;

嘿，这是一个有效的代码，我希望这对您有用。

Answer 4

$.ajax({
    url: 'http://some.working/url',
    type: "POST",
    data: $('#formContainer').serialize(),
    success: function (response) {
        console.log('Success', response);
    }, 
    error: function (response) {
        console.log('Error', response);
    }
});

The data can be produced in many ways for example
1. Using serialize() method as shown in the above example.
2. Using FormData():
   for example
   var data = new FormData($('#formContainer'));

In both of the above example, one thing compulsory is that your form 
must contain csrf field. which can be provided using any of the 
following methods:
<input type="hidden" name="_token" value="{{ csrf_token() }}" >
or 
{{ csrf_field() }}
or even more simply by just using 
@csrf

in some where in your form.

In case you are not using any form, you can create the data object by 
yourself like this
var data = {
   _token: '{{ csrf_token() }}',
   data1: 'Value1',
   data2: 'Value2',
   data3: 'Value2'
}

Laravel 5.7中ajax POST的最小工作示例

4 个答案: