Jquery 如何将Dropzone返回值附加到表单数据

我有一个Ajax，它通过laravel验证提交表单数据

$('body').on('click', '#submit', function(){
 var form_data = new FormData($('form')[1]);

 $.ajax({
     url:'/admin/products/',
     type:'POST',
     processData: false,
     contentType: false,
     dataType: 'json',
     data:form_data,
     success:function(data) {

     },
 });

然后在单独的Ajax url中上传/删除Dropzone文件

<div id="fileInput" class="dropzone">
    <input name="p_file" type="file" multiple />
</div>

Dropzone.autoDiscover = false;
new Dropzone('#fileInput', {
    // paramName: 'p_file',
    url: "/admin/products/upload",
    // autoProcessQueue: false,
    addRemoveLinks: true,

    init: function () {
        var myDropzone = this;
        myDropzone.on('success', function (file, serverResponse) {
            var fileuploded = file.previewElement.querySelector("[data-dz-name]");
            fileuploded.innerHTML = serverResponse;
        });

        myDropzone.on('removedfile', function(file, response) {
            var name = file.previewElement.querySelector('[data-dz-name]').innerHTML;
        });

        sending: function (file, xhr, formData) {
            // formData.append("name", $('#name').val());
        }
})

但我想将Dropzone值附加到第一个Ajax的form_数据中，而不是像文档中那样将其他form_数据附加到Dropzone中。

---

如何操作？

将dropzone文件分配到一个变量中，然后将其添加到FormData对象中

$('body').on('click', '#submit', function(){
var fileUpload = $('#fileInput').get(0).dropzone;
var files = fileUpload.files;
var form_data= new FormData();

// Looping over all files and add it to FormData object  
    for (var i = 0; i < files.length; i++) {
        form_data.append(files[i].name, files[i]);
    }

 $.ajax({
     url:'/admin/products/',
     type:'POST',
     processData: false,
     contentType: false,
     dataType: 'json',
     data:form_data,
     success:function(data) {

     },
 }); 

$('body').on('click', '#submit', function(){
var fileUpload = $('#fileInput').get(0).dropzone;
var files = fileUpload.files;
var form_data= new FormData();

// Looping over all files and add it to FormData object  
    for (var i = 0; i < files.length; i++) {
        form_data.append(files[i].name, files[i]);
    }

 $.ajax({
     url:'/admin/products/',
     type:'POST',
     processData: false,
     contentType: false,
     dataType: 'json',
     data:form_data,
     success:function(data) {

     },
 });