如何通过请求主机名设置代理与http-proxy-middleware和express?

问题描述 投票:1回答:1

我想用http-proxy-middleware和express配置代理。规则是主机名的映射,例如:

http://123.com   >>  http://localhost:3000/123
http://456.com   >>  http://localhost:3000/abc

我试过这样的:

import express from 'express';
import http from 'http';
import proxy from 'http-proxy-middleware';

const app = express();

app.use( async function (req, res) {  
  let direction = 'http://localhost:3000';
  console.log('hostname: ', req.hostname);
  console.log('originalUrl: ', req.originalUrl);
  if (req.hostname == '123.com') {
    direction = `http://localhost:3000/123${req.originalUrl}`;
  }
  if (req.hostname == '456.com') {
    direction = `http://localhost:3000/abc${req.originalUrl}`;
  }

  return await proxy({ target: direction, changeOrigin: false })
});

const server = http.createServer(app);
app.set('port', '127.0.0.1');
server.listen(9999, '0.0.0.0');

但它不起作用。

node.js http-proxy http-proxy-middleware
1个回答
2
投票

您需要考虑以下几点:

  • http-proxy-middleware模块不返回承诺,而是返回一个快速中间件。
  • 您可以使用自定义过滤器来决定是否代理
  • 您需要添加pathRewrite选项,以便根据当前主机名重写URL。 另一种选择是使用路由器选项 - >查看https://github.com/chimurai/http-proxy-middleware#options以获取更多信息

我写了一个快速的快速应用程序来测试这个(注意我用localwebapplocalwebapp2指向127.0.0.1覆盖我的主机文件)它似乎工作正常:

const express = require('express')
const proxy = require('http-proxy-middleware')

const app = express();
const filter = (pathname, req) => {
    if (req.hostname == 'localwebapp' || req.hostname == 'localwebapp2') {
        return true;
    }
    return false;
};

app.get('/123*', (req, res) => {
    res.send(`matched 123* route: ${req.path}`);
})

app.get('/abc*', (req, res) => {
    res.send(`matched abc* route: ${req.path}`);
})

app.get('/test', (req, res) => {
    res.send("matched non proxied route '/test'");
})

const apiProxy = proxy(filter, {
    target: 'http://localhost:3000', logLevel: 'debug',
    changeOrigin: true,
    pathRewrite: function (path, req) {
        if (req.hostname == 'localwebapp') {
            return `/123${req.originalUrl}`;
        }
        if (req.hostname == 'localwebapp2') {
            return `/abc${req.originalUrl}`;
        }
        return path;
    }
})
app.use(apiProxy)
app.listen(3000);
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