Node.js：直接用WHATWG URL解析req.url报错，需从旧URL API迁移

环境信息：Node.js v18.15.0、Ubuntu 22.04、NPM 9.5.0

在创建Node.js HTTP服务器路由时，直接将req.url传入WHATWG URL构造函数会抛出无效URL错误，报错代码如下：

const httpServer = http.createServer((req, res) => {
    res.writeHead(200, {'Content-Type': 'text/html'});
    const requestURL = new URL(req.url);
    const requestHostname = requestURL.hostname;
    const requestPathname = requestURL.pathname;
    const requestParams = requestURL.searchParams;
    console.log(requestPathname);
});

对应的报错信息：

node:internal/url:566
  throw new ERR_INVALID_URL(input);
  ^

TypeError [ERR_INVALID_URL]: Invalid URL
    at new NodeError (node:internal/errors:399:5)
    at URL.onParseError (node:internal/url:566:9)
    at new URL (node:internal/url:646:5)
    at Server.<anonymous> (/var/www/newsat11.live/index.js:26:24)
    at Server.emit (node:events:513:28)
    at parserOnIncoming (node:_http_server:1072:12)
    at HTTPParser.parserOnHeadersComplete (node:_http_common:119:17) {
  input: '/',
  code: 'ERR_INVALID_URL'
}

Node.js v18.15.0

而使用Node.js URL模块的传统API url.parse(req.url) 则可正常运行：

const httpServer = http.createServer((req, res) => {
    res.writeHead(200, {'Content-Type': 'text/html'});
    const requestURL = url.parse(req.url);
    const requestHostname = requestURL.hostname;
    const requestPathname = requestURL.pathname;
    const requestParams = requestURL.searchParams;
    console.log(requestPathname); // logs "/" as the pathname for localhost
});

问题原因 #

WHATWG URL构造函数要求传入完整的URL字符串（必须包含协议、主机等核心部分），但req.url仅包含请求的路径和查询参数（比如/或/api?foo=bar），并非完整URL，因此会触发无效URL错误。而传统的url.parse方法支持解析相对路径，所以可以正常工作。

解决方案 #

通过拼接协议和请求头中的host字段，构造完整的URL后再传入WHATWG URL构造函数。示例代码如下：

const httpServer = http.createServer((req, res) => {
    res.writeHead(200, {'Content-Type': 'text/html'});
    // 拼接完整URL：协议（这里用http，若为HTTPS则替换为https）+ host + req.url
    const fullURL = `http://${req.headers.host}${req.url}`;
    const requestURL = new URL(fullURL);
    const requestHostname = requestURL.hostname;
    const requestPathname = requestURL.pathname;
    const requestParams = requestURL.searchParams;
    console.log(requestPathname);
});

如果服务器运行在HTTPS环境下，只需将协议部分改为https://即可：

const fullURL = `https://${req.headers.host}${req.url}`;

内容的提问来源于stack exchange，提问作者Newsat11