Question

如何将我的sitemap.xml（仅此文件）的HTTP标头“ X-Robots-Tag”设置为“ noindex，跟随” express？

sitemap.xml位于我的应用程序的根目录。

这是我的exress配置的摘要：

express / development.js

const express = require('express')
const webpack = require('webpack')
const cors = require('cors')
const webpackDevMiddleware = require('webpack-dev-middleware')
const webpackHotMiddleware = require('webpack-hot-middleware')
const webpackHotServerMiddleware = require('webpack-hot-server-middleware')
const config = require('./../webpack/webpack.development.config.js')

const app = express()
const compiler = webpack(config)
const PORT = process.env.PORT || 3000

app.use(cors())
app.use(express.static('build'))
app.use(express.static('public'))

...

我已经尝试过：

... 

app.get('/sitemap.xml', res => {
  res.setHeader('X-Robots-Tag', 'noindex, follow')
})

还有：

app.get('/sitemap.xml', res => {
  res.set('X-Robots-Tag', 'noindex, follow')
})

此外，如何查看标头是否已更新。

我已经尝试过：$curl -v localhost:3000/sitemap.xml

但是标头中没有'X-Robots-Tag'，所以我认为它不起作用。

Answer 1

如docs中所述，静态中间件接受一个options参数，您可以在其中设置const fs = require('fs'); const path = require('path'); const http = require('http'); const express = require('express'); const publicDirectoryPath = path.join(__dirname, './public'); // the following creates and a public directory in the script directory and populates it with some files (for the sake of this demo). // create the public directory if it doesn't exist if (!fs.existsSync(publicDirectoryPath)) { fs.mkdirSync(publicDirectoryPath); } // create a sample sitemap.xml (for the sake of this demo) const siteMapXml = `<?xml version="1.0" encoding="UTF-8"?> <urlset xmlns="http://www.sitemaps.org/schemas/sitemap/0.9"> <url> <loc>http://www.example.com/</loc> <lastmod>2005-01-01</lastmod> <changefreq>monthly</changefreq> <priority>0.8</priority> </url> </urlset>`; fs.writeFileSync(path.join(publicDirectoryPath, './sitemap.xml'), siteMapXml); // create a sample index.html const indexHtml = `<!DOCTYPE html> <html lang="en"> <meta charset="utf-8"> <title>Demo</title> <p>It works!</p> </html>`; fs.writeFileSync(path.join(publicDirectoryPath, './index.html'), indexHtml); // setup and start the express server const app = express(); const PORT = process.env.PORT || 3000; app.use( express.static(path.join(__dirname, "public"), { setHeaders: (res, path) => { if (path.endsWith("sitemap.xml")) { res.setHeader("X-Robots-Tag", "noindex, follow"); } } }) ); app.listen(PORT); // check response headers const makeGetRequestAndPrintHeaders = (path) => { const options = { hostname: 'localhost', port: PORT, path, method: 'GET' }; const req = http.request(options, (res) => { console.log(`GET ${path}`); console.log(JSON.stringify(res.headers, null, 2)); }); req.end(); } makeGetRequestAndPrintHeaders('/index.html'); makeGetRequestAndPrintHeaders('/sitemap.xml'); // example output /* GET /index.html { "x-powered-by": "Express", "accept-ranges": "bytes", "cache-control": "public, max-age=0", "last-modified": "Wed, 13 Feb 2019 14:28:26 GMT", "etag": "W/\"6b-168e74243aa\"", "content-type": "text/html; charset=UTF-8", "content-length": "107", "date": "Wed, 13 Feb 2019 14:28:26 GMT", "connection": "close" } GET /sitemap.xml { "x-powered-by": "Express", "x-robots-tag": "noindex, follow", "accept-ranges": "bytes", "cache-control": "public, max-age=0", "last-modified": "Wed, 13 Feb 2019 14:28:26 GMT", "etag": "W/\"113-168e74243a9\"", "content-type": "text/xml; charset=UTF-8", "content-length": "275", "date": "Wed, 13 Feb 2019 14:28:26 GMT", "connection": "close" } */函数以基于路径修改响应标头：

{{1}}

Answer 2

TLDR

您实际上是在为请求设置标题，下面的代码应该可以工作：

app.get('/sitemap.xml', (req, res) => {
  res.setHeader('X-Robots-Tag', 'noindex, follow')
  res.sendFile(`${__dirname}/sitemap.xml`)
})

对于app.get，回调函数具有三个参数：

app.get('/sitemap.xml', (request, response, nextHandler) => {})

request是来自客户端的请求对象。

response是您将从服务器发送的响应，这是您实际要更改的内容。

{nextHandler通过调用（nextHandler()）将所有内容传递给您的下一个请求处理程序

如何使用Express

2 个答案:

TLDR