jquery - 如何在mvc中使用防伪 token 发出ajax请求

Question

我对 MVC 项目的以下详细信息有疑问。

当我尝试使用 jquery ajax 请求与加载面板(如旋转 gif(甚至文本))时，我收到错误，从 fiddler 观察到

The required anti-forgery form field "__RequestVerificationToken" is not present.

如果我评论 [ValidateAntiForgeryToken] attribute在 POST 操作方法并使用加载面板时，它工作正常。我想知道为什么会收到此错误。

我什至使用了序列化的查询字符串

__RequestVerificationToken= $('input[name="__RequestVerificationToken"').val()

我仍然收到错误

The anti-forgery token could not be decrypted. If this application is hosted by a Web Farm or cluster, ensure that all machines are running the same version of ASP.NET Web Pages and that the <machineKey> configuration specifies explicit encryption and validation keys.

AutoGenerate 不能在集群中使用

我应该使用什么？

这里更新了问题代码

var token = $('input[name="__RequestVerificationToken"]').val();
$('#submitaddress').click(function subaddr(event) {
    event.preventDefault();
    event.stopPropagation();
  //$('#addAddress').html('<img src="/img/animated-overlay.gif"> Sending...');
   // $('#addAddress').blur();
    //  $(this).bl
    if ($('#Jobid').val()!="") {
        $('#TransportJobId').val(parseInt($('#Jobid').val()));
        $.ajax(
              {
                  url: '/TransportJobAddress/create',
                  type: 'POST',
                  data: "__RequestVerificationToken=" + token + "" + $('form[action="/TransportJobAddress/Create"]').serialize(),
                  success: function poste(data, textStatus, jqXHR) { $('#addAddress').html(data); return false; },
                  error: function err(jqXHR, textStatus, errorThrown) { alert('error at address :' + errorThrown); }
              });
    }
    else {
        var transportid = 2;
        $.ajax({
            url: '/TransportJob/create',
            type: 'POST',
            data: "__RequestVerificationToken=" + token + "" + $('form[action="/TransportJob/Create"]').serialize(),
            success: function sfn(data, textStatus, jqXHR) {
                transportid = parseInt(data);
                $('#Jobid').val(data);
               // alert('inserted id :' + data);
                $('#TransportJobId').val((transportid));
                $.ajax(
         {

             url: '/TransportJobAddress/create',
             type: 'POST',
             //beforeSend: function myintserver(xhr){  
             //        $('#addAddress').html('<div id="temp_load" style="text-align:center">please wait ...</div>');
             //}, 
             data: "__RequestVerificationToken=" + token + "" + $('form[action="/TransportJobAddress/Create"]').serialize(),
             success: function poste(data, textStatus, jqXHR) {
                 $('#addAddress').html(data);

             },
             error: function err(jqXHR, textStatus, errorThrown) {
                 alert('error at address :' + errorThrown);
             }

         });
            },
            error: function myfunction(jqXHR, textStatus, errorThrown) {
                alert("error at transport :" + jqXHR.textStatus);
            },
            complete: function completefunc() {
              //  alert('ajax completed all requests');
                return false;
            }

        });
    }
});

表单标签

<form action="/TransportJob/Create" method="post"><input     name="__RequestVerificationToken" type="hidden"   value="ydYSei0_RfyBf619dQrhDwwoCM7OwWkJQQEMNvNdAkefiFfYvRQ0MJYYu0zkktNxlJk_y1ZJO9-yb-  COap8mqd0cvh8cDYYik4HJ0pZXTgE1" />   

运输工作同一页面上的表单标签 2

<form action="/TransportJobAddress/Create" method="post" novalidate="novalidate"><input name="__RequestVerificationToken" type="hidden"    value="Np2vUZJPk1TJlv846oPSU6hg4SjMHRcCk1CacaqZbpHOg8WbV4GZv06noRDl7F_iT9qQf3BIXo3n9wGW68sU mki7g3-ku_BSHBDN-g2aaKc1"> 

Answer 1

我通常不会手动将其添加到每个请求中，而是这样做:

var token = $('input[name="__RequestVerificationToken"]').val();
$.ajaxPrefilter(function (options, originalOptions) {
  if (options.type.toUpperCase() == "POST") {
    options.data = $.param($.extend(originalOptions.data, { __RequestVerificationToken: token }));
  }
});

这会自动将您的 token 添加到您执行的任何 ajax POST 中。