Question

我从C＃发送一个POST WebRequest以及一个JSON对象数据，并希望在Node.js服务器中接收它，如下所示：

var express = require('express');
var app = express.createServer();

app.configure(function(){
  app.use(express.bodyParser());
});
app.post('/ReceiveJSON', function(req, res){
  //Suppose I sent this data: {"a":2,"b":3}

  //Now how to extract this data from req here?  

  //console.log("req a:"+req.body.a);//outputs 'undefined'
  //console.log("req body:"+req.body);//outputs '[object object]'


  res.send("ok");
});

app.listen(3000);
console.log('listening to http://localhost:3000');

此外，通过以下方法调用POST WebRequest的C＃结尾：

public string TestPOSTWebRequest(string url,object data)
{
    try
    {
        string reponseData = string.Empty;

        var webRequest = System.Net.WebRequest.Create(url) as HttpWebRequest;
        if (webRequest != null)
        {
            webRequest.Method = "POST";
            webRequest.ServicePoint.Expect100Continue = false;
            webRequest.Timeout = 20000;

            webRequest.ContentType = "application/json; charset=utf-8";
            DataContractJsonSerializer ser = new DataContractJsonSerializer(data.GetType());
            MemoryStream ms = new MemoryStream();
            ser.WriteObject(ms, data);
            String json = Encoding.UTF8.GetString(ms.ToArray());
            StreamWriter writer = new StreamWriter(webRequest.GetRequestStream());
            writer.Write(json);
        }

        var resp = (HttpWebResponse)webRequest.GetResponse();
        Stream resStream = resp.GetResponseStream();
        StreamReader reader = new StreamReader(resStream);
        reponseData = reader.ReadToEnd();

        return reponseData;
    }
    catch (Exception x)
    {
        throw x;
    }
}

方法调用：

TestPOSTWebRequest("http://localhost:3000/ReceiveJSON", new TestJSONType {a = 2, b = 3});

如何在上面的Node.js代码中解析来自请求对象的JSON数据？

Answer 1

请求必须与以下内容一起发送： content-type：“application / json; charset = utf-8”

否则bodyParser会将您的对象踢为另一个对象中的键：）

Answer 2

bodyParser会自动为您执行此操作，只需执行console.log(req.body)

编辑：您的代码错误，因为您首先在bodyParser和其他所有内容之前包含app.router（）。那很糟。您甚至不应该包含app.router（），Express会自动为您执行此操作。以下是代码的外观：

var express = require('express');
var app = express.createServer();

app.configure(function(){
  app.use(express.bodyParser());
});

app.post('/ReceiveJSON', function(req, res){
  console.log(req.body);
  res.send("ok");
});

app.listen(3000);
console.log('listening to http://localhost:3000');

你可以使用Mike的漂亮的Request模块测试这个，通过发送带有这些参数的POST请求：

var request = require('request');
request.post({
  url: 'http://localhost:3000/ReceiveJSON',
  headers: {
    'Content-Type': 'application/json'
  },
  body: JSON.stringify({
    a: 1,
    b: 2,
    c: 3
  })
}, function(error, response, body){
  console.log(body);
});

更新：使用body-parser代表快递4 +。

如何在express node.js POST请求中接收JSON？

2 个答案: