使用node.js实现apache功能

1、先实现在url中输入文件路径能展示对应文件内容功能

const http = require('http')
const fs = require('fs')

const server = http.createServer()

const wwwDir = '/Users/lianglanlan/Desktop/code/study/node/www'

server.on('request', (req, res) => {
    const url = req.url

    let filePath = '/index.html'

    if (url !== '/') {
        filePath = url
    }
    fs.readFile(wwwDir + filePath, (err, data) => {
        if (err) {
            console.log(err)
            return res.end('404 Not Found')
        }
        res.end(data)
    })
})

server.listen(3010, () => {
    console.log('running...')
})

2、展示页面template.html

用谷歌浏览器打开文件夹,复制整个html,删除无用js

简单写出这样的html页面

3、修改requset事件中url对应的展示页面为template.html

server.on('request', (req, res) => {
    fs.readFile('./template.html', (err, data) => {
        if (err) {
            return res.end('404 Not Found')
        }
        res.end(data)
    })
})
原文地址:https://www.cnblogs.com/lianglanlan/p/12196099.html