Express.js发布请求“原产地不允许”错误

Question

我想发布一个json数据，其中包含一组键和值从JavaScript到Express.JS。按照其他帖子的建议添加了cors，并且还添加了res.header("Access-Control-Allow-Origin", "*");。
当我测试Chrome邮递员应用程序的POST请求时，它工作得很好。但是，当我通过Chrome浏览器通过本地测试网页发送请求，控制台输出
XMLHttpRequest cannot load http://localhost:8080/submit_project_request. No 'Access-Control-Allow-Origin' header is present on the requested resource. Origin 'null' is therefore not allowed access. The response had HTTP status code 400.

我对Express.JS代码：

var http = require("http"); 
var express = require("express"); 
var xmlbuilder = require("xmlbuilder"); 
var bodyParser = require("body-parser"); 
var cors = require("cors"); 

var app = express(); 
app.use(cors()); 
var port = process.env.PORT || 8080; 

app.use(bodyParser.urlencoded({ 
    extended: true 
})); 

app.use(bodyParser.json()); 

app.use(function (req, res, next) { 
    res.header("Access-Control-Allow-Origin", "*"); 
    res.header("Access-Control-Allow-Headers", "Origin, X-Requested-With, Content-Type, Accept"); 
    next(); 
}); 

app.get('/submit_project_request', function (req, res) { 
    var client_org = req.parm("client_org"); 
}); 

app.post('/submit_project_request', function (req, res, next) { 
    res.header("Access-Control-Allow-Origin", "*"); 
    res.header("Access-Control-Allow-Headers", "X-Requested-With"); 
    next(); 
    console.log(req.body.client_org); 
    res.contentType("text"); 
    res.send("data received"); 
}); 

app.listen(8080); 

请随时提出任何改善。

Answer 1

看来你正在使用file：//打开你的网页。使用一台服务器，让你看到类似本地主机提供给您的html页面：8000/yourpage.html

而且，你并不需要以下中间件（如果你想配置CORS模块然后使用选项对象像app.use(cors(options))查看可用选项here）：

app.use(function (req, res, next) { 
    res.header("Access-Control-Allow-Origin", "*"); 
    res.header("Access-Control-Allow-Headers", "Origin, X-Requested-With, Content-Type, Accept"); 
    next(); 
}); 

你也可能要删除您submit_project_request路线如下：

res.header("Access-Control-Allow-Origin", "*"); 
    res.header("Access-Control-Allow-Headers", "X-Requested-With"); 
    next(); 

调用next()这样是不推荐。在你的代码中，它基本上意味着在这里发送404。如果你打算这样做，那么使用return next();