Question

我想从python服务器发送一个结果作为对客户端JavaScript的响应。我的JavaScript代码在这里：

document.addEventListener('DOMContentLoaded', function() {
chrome.tabs.getSelected(null, function(tab) {
  d = document;
  var xhr = new XMLHttpRequest();
  var f = d.createElement('form');
  f.action = 'http://127.0.0.1:5000/Get%20Form/';
  f.method = 'post';
  var i = d.createElement('input');
  i.type = 'hidden';
  i.name = 'url';
  i.value = tab.url;
  f.appendChild(i);
  d.body.appendChild(f);
  f.submit();
});
}, false);

还有我的python服务器代码如下：

from django.http import HttpResponse
import json
from flask import Flask, flash, redirect, url_for, request, 
  render_template,jsonify
app = Flask(__name__)
user=""@app.route('/Get Form/',methods = ['POST'])
def GetForm():`
  request.method == 'POST'
  url=request.form['url']  
  print url
  data={}
  data['url'] = url 
  data['result'] = 'safe'
  response = HttpResponse(json.dumps(data), content_type = "application/json")      
  print response
  return

if (__name__ == '__main__'):
app.run(debug=True,host='0.0.0.0', use_reloader=False)

任何正文帮助我将JSON对象响应发送到客户端JavaScript

Answer 1

您可以在问题

的代码中使用FormData() XMLHttpRequest()变量xhr

  var xhr = new XMLHttpRequest();
  xhr.open("POST", "http://127.0.0.1:5000/Get%20Form/");
  xhr.onload = function() {
    console.log(xhr.responseText) // do stuff with response from server
  }
  xhr.onerror = function(err) {
    console.error(err)
  }
  var fd = new FormData();
  fd.append("url", tab.url);
  xhr.send(fd);

提交表单后如何从服务器发送响应？

1 个答案: