如何在Node.js中提取没有HTTP的完整URL?
我想提取除HTTP之外的完整URL。我用过和 这是我的密码如何在Node.js中提取没有HTTP的完整URL?,node.js,url,Node.js,Url,我想提取除HTTP之外的完整URL。我用过和 这是我的密码 var url = require('url'); var domain=require('domain.js'); var url_parts = 'http://static01.nyt.com/images/2014/11/17/business/billboardjump/billboardjump-master675.jpg'; var website=domain(url.parse(url_parts)); var
var url = require('url');
var domain=require('domain.js');
var url_parts = 'http://static01.nyt.com/images/2014/11/17/business/billboardjump/billboardjump-master675.jpg';
var website=domain(url.parse(url_parts));
var querystring = (url.parse(url_parts, true)).path;
console.log(website+querystring);
但只有我得到了
'nyt.com/images/2014/11/17/business/billboardjump/billboardjump-master675.jpg'
而不是
'static01.nyt.com/images/2014/11/17/business/billboardjump/billboardjump-master675.jpg'
提前谢谢 只需计算协议长度并将其删除:
var u = 'http://static01.nyt.com/images/2014/11/17/business/billboardjump/billboardjump-master675.jpg';
var protocol = url.parse(u).protocol;
console.log(u.slice((protocol + '//').length));
如果最后进行字符串切片,为什么要使用URL解析器呢?因为它已经可用,您不需要自己解析协议部分
const urlString = "https://stackoverflow.com/questions/26987567/how-to-extract-complete-url-without-http-in-node-js"
const [protocol, urlWithoutProtocol] = urlString.split('://');
console.log(protocol, urlWithoutProtocol)