通过ajax发送FormData对象和一个附加参数 [英] Send FormData object AND an additional parameter via ajax

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本文介绍了通过ajax发送FormData对象和一个附加参数的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我设法发送了一个FormData对象,如下所示:

I have managed to send a FormData object like so:

var formData = new FormData();
formData.append('file', this.files[0]);
$.ajax({
   url: urlUploadProductsFile,
   type: 'POST',
   data: formData,
   cache: false,
   contentType: false,
   processData: false
}, 'json');

现在我要做的是添加一个附加的CustomerId以发送到服务器.以下内容不起作用:

Now what I want to do is add an additional CustomerId to send to the server. The following won't work:

var formData = new FormData();
formData.append('file', this.files[0]);
$.ajax({
   url: urlUploadProductsFile,
   type: 'POST',
   data: { "file": formData, "CustomerId": 2 },
   cache: false,
   contentType: false,
   processData: false
}, 'json');

我还尝试了以下变体:

data: { "file": formData, "CustomerId": 2 }, processData: true

data: JSON.stringify({ "file": formData, "CustomerId": 2 })

data: { "file": JSON.stringify(formData), "CustomerId": 2 }

data: { file: formData, CustomerId: 2 }

任何帮助表示赞赏.

推荐答案

尝试:

var formData = new FormData();
formData.append('file', this.files[0]);
formData.append('CustomerId', 2);

/*
 note:: appending in form Data will give "csrf token mismatch error". 
 so better you make a input feild of type hidden with name = CustomerId 
 and value =  2 
*/ 

$.ajax({
   url: urlUploadProductsFile,
   type: 'POST',
   data: formData,
   cache: false,
   contentType: false,
   processData: false
}, 'json');

这篇关于通过ajax发送FormData对象和一个附加参数的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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