如何确定 XMLHttpRequest.send() 是否有效

how to find out if XMLHttpRequest.send() worked(如何确定 XMLHttpRequest.send() 是否有效)

问题描述

我正在使用 XMLHttpRequest 将文件从 javascript 代码发送到 django 视图.我需要检测该文件是否已被发送或如果发生错误.我使用 jquery 编写以下 javascript.

I am using XMLHttpRequest to send a file from javascript code to a django view.I need to detect,whether the file has been sent or if some error occurred.I used jquery to write the following javascript.

理想情况下，我想向用户显示文件未上传的错误消息.有没有办法在 javascript 中做到这一点?

Ideally I would like to show the user an error message that the file was not uploaded.Is there some way to do this in javascript?

我试图通过从 django view 返回 success/failure 消息来做到这一点，将 success/failed 消息 作为 json 并从 django 视图 发回序列化的 json.为此，我制作了 xhr.open() non-asynchronous.我试图打印 xmlhttpRequest 对象的 responseText . console.log(xhr.responseText) 显示

I tried to do this by returning a success/failure message from django view , putting the success/failed message as json and sending back the serialized json from the django view.For this,I made the xhr.open() non-asynchronous. I tried to print the xmlhttpRequest object's responseText .The console.log(xhr.responseText) shows

response= {"message": "success"}

我想知道的是，这是否是正确的方法.在许多文章中，我发现警告

What I am wondering is,whether this is the proper way to do this.In many articles,I found the warning that

不推荐使用 async=false

Using async=false is not recommended

那么，有没有什么办法可以在保持xhr.open()异步的同时查出文件是否已经发送?

So,is there any way to find out whether the file has been sent,while keeping xhr.open() asynchronous?

$(document).ready(function(){
   $(document).on('change', '#fselect', function(e){
            e.preventDefault();
            sendFile();
        });
});

function sendFile(){
   var form = $('#fileform').get(0);
   var formData = new FormData(form);
   var file = $('#fselect').get(0).files[0];
   var xhr = new XMLHttpRequest();
   formData.append('myfile', file);
   xhr.open('POST', 'uploadfile/', false);
   xhr.send(formData);
   console.log('response=',xhr.responseText);
}

我的 django 视图从表单数据中提取文件并写入目标文件夹.

My django view extracts file from form data and writes to a destination folder.

def store_uploaded_file(request):
   message='failed'
   to_return = {}
   if  (request.method == 'POST'):          
      if request.FILES.has_key('myfile'):
         file = request.FILES['myfile']
         with open('/uploadpath/%s' % file.name, 'wb+') as dest:
            for chunk in file.chunks():
               dest.write(chunk)
               message="success"
   to_return['message']= message
   serialized = simplejson.dumps(to_return)
   if store_message == "success":
      return HttpResponse(serialized, mimetype="application/json")
   else:
      return HttpResponseServerError(serialized, mimetype="application/json")

我在 @FabrícioMatté 的帮助下完成了这项工作

I got this working with the help of @FabrícioMatté

xhr.onreadystatechange=function(){
       if (xhr.readyState==4 && xhr.status==200){
          console.log('xhr.readyState=',xhr.readyState);
          console.log('xhr.status=',xhr.status);
          console.log('response=',xhr.responseText);

          var data = $.parseJSON(xhr.responseText);
          var uploadResult = data['message']
          console.log('uploadResult=',uploadResult);

          if (uploadResult=='failure'){
             console.log('failed to upload file');
             displayError('failed to upload');
          }else if (uploadResult=='success'){
             console.log('successfully uploaded file');
          }
       }
    }

推荐答案

XMLHttpRequest 对象包含 status 和 readyState 属性，您可以在 xhr.onreadystatechange 事件中进行测试以检查您的请求是否成功.

XMLHttpRequest objects contain the status and readyState properties, which you can test in the xhr.onreadystatechange event to check if your request was successful.

这篇关于如何确定 XMLHttpRequest.send() 是否有效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编程学习网！