node.js 用于 Express 中路由匹配的正则表达式
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Regex for route matching in Express
提问by Jonathan Ong
I'm not very good with regular expressions, so I want to make sure I'm doing this correctly. Let's say I have two very similar routes, /discussion/:slug/and /page/:slug/. I want to create a route that matches both these pages.
我不太擅长正则表达式，所以我想确保我正确地做到了这一点。假设我有两条非常相似的路线，/discussion/:slug/并且/page/:slug/. 我想创建一个匹配这两个页面的路线。
app.get('/[discussion|page]/:slug', function(req, res, next) {
...enter code here...
})
Is this the correct way to do it? Right now I'm just creating two separate routes.
这是正确的方法吗？现在我只是在创建两条独立的路线。
someFunction = function(req, res, next) {..}
app.get('/discussion/:slug', someFunction)
app.get('/page/:slug', someFunction)
回答by Jonathan Ong
app.get('/:type(discussion|page)/:id', ...)works
app.get('/:type(discussion|page)/:id', ...)作品
回答by Peter Lyons
You should use a literal javascript regular expression object, not a string, and @sarnold is correct that you want parens for alternation. Square brackets are for character classes.
您应该使用文字 javascript 正则表达式对象，而不是字符串，@sarnold 是正确的，您希望使用括号进行交替。方括号用于字符类。
const express = require("express");
const app = express.createServer();
app.get(/^\/(discussion|page)\/(.+)/, function (req, res, next) {
res.write(req.params[0]); //This has "discussion" or "page"
res.write(req.params[1]); //This has the slug
res.end();
});
app.listen(9060);
The (.+)means a slug of at least 1 character must be present or this route will not match. Use (.*)if you want it to match an empty slug as well.
这(.+)意味着必须存在至少 1 个字符的 slug，否则此路由将不匹配。(.*)如果您希望它也与空弹头相匹配，请使用。
