假设我的示例URL是
http://example.com/one/two
我说我有以下路线
app.get('/one/two', function (req, res) {
var url = req.url;
}
url的值是/one/two。
如何在Express中获得完整的URL ? 例如,在上面的情况下,我想收到http://example.com/one/two。
假设我的示例URL是
http://example.com/one/two
我说我有以下路线
app.get('/one/two', function (req, res) {
var url = req.url;
}
url的值是/one/two。
如何在Express中获得完整的URL ? 例如,在上面的情况下,我想收到http://example.com/one/two。
当前回答
var full_address = req.protocol + "://" + req.headers.host + req.originalUrl;
or
var full_address = req.protocol + "://" + req.headers.host + req.baseUrl;
其他回答
谢谢大家提供这些信息。这是非常烦人的。
把这个添加到你的代码中,你就再也不用考虑它了:
var app = express();
app.all("*", function (req, res, next) { // runs on ALL requests
req.fullUrl = req.protocol + '://' + req.get('host') + req.originalUrl
next()
})
您还可以在那里执行或设置其他操作,例如将日志记录到控制台。
2021年
上面的答案工作得很好,但不是首选的文档,因为url。解析现在是遗留的,所以我建议你使用新的URL()函数,如果你想获得更多的控制URL。
表达方式
您可以从下面的代码获得完整的URL。
`${req.protocol}://${req.get('host')}${req.originalUrl}`
示例URL: http://localhost:5000/a/b/c?d=true&e=true#f=false
固定属性(你将在所有路线上得到相同的结果)
req.protocol: http
req.hostname: localhost
req.get('Host'): localhost:5000
req.originalUrl: /a/b/c?d=true&e=true
req.query: { d: 'true', e: 'true' }
非固定属性(将在每条路由中改变,因为它由express自己控制)
路线:/
req.baseUrl: <blank>
req.url: /a/b/c?d=true&e=true
req.path: /a/b/c
路线/
req.baseUrl: /a
req.url: /b/c?d=true&e=true
req.path: /b/c
文档:http://expressjs.com/en/api.html req.baseUrl
URL封装方式
在URL函数中,您将在每个路由中得到相同的结果,因此属性总是固定的。
属性
const url = new URL(`${req.protocol}://${req.get('host')}${req.originalUrl}`);
console.log(url)
您将得到如下结果。我根据图像改变了属性的顺序,这样它就可以匹配图像流。
URL {
href: 'http://localhost:5000/a/b/c?d=true&e=true',
protocol: 'http:',
username: '',
password: '',
hostname: 'localhost',
port: '5000',
host: 'localhost:5000',
origin: 'http://localhost:5000',
pathname: '/a/b/c',
search: '?d=true&e=true',
searchParams: URLSearchParams { 'd' => 'true', 'e' => 'true' },
hash: ''
}
注意:散列不能发送到服务器,因为它在服务器中被视为片段,但你会在客户端浏览器中获得。
文档:https://nodejs.org/api/url.html url_new_url_input_base
The protocol is available as req.protocol. docs here Before express 3.0, the protocol you can assume to be http unless you see that req.get('X-Forwarded-Protocol') is set and has the value https, in which case you know that's your protocol The host comes from req.get('host') as Gopal has indicated Hopefully you don't need a non-standard port in your URLs, but if you did need to know it you'd have it in your application state because it's whatever you passed to app.listen at server startup time. However, in the case of local development on a non-standard port, Chrome seems to include the port in the host header so req.get('host') returns localhost:3000, for example. So at least for the cases of a production site on a standard port and browsing directly to your express app (without reverse proxy), the host header seems to do the right thing regarding the port in the URL. The path comes from req.originalUrl (thanks @pgrassant). Note this DOES include the query string. docs here on req.url and req.originalUrl. Depending on what you intend to do with the URL, originalUrl may or may not be the correct value as compared to req.url.
将这些组合在一起以重建绝对URL。
var fullUrl = req.protocol + '://' + req.get('host') + req.originalUrl;
你可以使用node.js API来获取url,并将来自express的信息传递给URL.format(),而不是自己将这些东西连接在一起。
例子:
var url = require('url');
function fullUrl(req) {
return url.format({
protocol: req.protocol,
host: req.get('host'),
pathname: req.originalUrl
});
}
你需要使用req.headers.host + req.url来构造它。当然,如果你是在不同的端口,你得到的想法;-)