TypeError：对象不是一个函数node.js我可以，我似乎得到的地方我做错了

var fs = require("fs"); var config = JSON.parse(fs.readFileSync("serverConfig.json")); var port = config.port; var host = config.host; var express = require("express"); var app = express(); 

我正在尝试Node.js我想要实现的是打印链接引用Twitter的地方，如果他们可以跟随两个用户，如果他们在url栏中input正确的用户ID

 //app.use(express.static(__dirname + "/public")); app.get('/', function(req, res){ res.send("hello"); }); app.get('/hello/:text', function(request, responce){ responce.send("Hello " + request.params.text); }); //user objects with user ID 1 and 2 var users = { "1" : { "name": "Lindo", "twitter": "lynndor_D" }, "2" : { "name": "Xolani", "twitter": "mcxolani" } }; app.get('/user/:id', function(request, responce){ var user = users[request.params.id]; if(user){ responce("Hello" + user.name); responce("<a href= http://twitter.com" +user.twitter +"'>follow" + user.name + "on twitter</a>"); }else{ responce("Sorry! user not found", 404);// 404 is an http status code for "page not found" } }); 

app.listen（port，host）;

  enter code here 

Answer 1

你正在使用responce作为一个函数，它是一个对象。 您需要调用方法。发送给用户的响应。

像这样的东西：

 app.get('/user/:id', function(request, responce){ var user = users[request.params.id]; if(user){ responce.write("Hello" + user.name); responce.write("<a href= http://twitter.com" +user.twitter +"'>follow" + user.name + "on twitter</a>"); responce.end(); }else{ responce.end("Sorry! user not found", 404);// 404 is an http status code for "page not found" } });