js匹配获取一级域名后的字符串

Question

研究了好一会，没搞定(lll￢ω￢)

var imgarr = ['https://www.baidu.com/123.jpg','https://www.baidu.com/abc.jpg','https://www.baidu.com/nana.jpg']

输出结果是字符串

var imgarr = '/123.jpg','/abc.jpg','/nana.jpg' 

我不会匹配一级域名，求指教

Answer 1

如果不考虑兼容性，这可能是最好的方法：URL

var imgarr = ['https://www.baidu.com/123.jpg', 'https://www.baidu.com/abc.jpg', 'https://www.baidu.com/nana.jpg'];
console.log(imgarr.map(v => (new URL(v)).pathname));//["/123.jpg", "/abc.jpg", "/nana.jpg"]

Answer 2

var imgarr = ('https://www.baidu.com/index/123.jpg'.match(/\b\/.*/g) || [])[0]

Answer 3

window.location.pathname

Answer 4

一个比较靠谱的一级域名匹配表达式/((https?|ftp|file):\/\/)?[-A-Za-z0-9+&@#\/%?=~_|!:,.;]+\.[-A-Za-z+]+\/+/g。

let imgarr = ['https://www.baidu.com/123.jpg','https://www.baidu.com/abc.jpg','https://www.baidu.com/nana.jpg'],
    reg = /((https?|ftp|file):\/\/)?[-A-Za-z0-9+&@#\/%?=~_|!:,.;]+\.[-A-Za-z+]+\/+/g;

imgarr = imgarr.map(item => item.replace(reg, '/'))

Answer 5

'http://www.baidu.com/123.jpg'.replace(/https?\:\/\/[a-z0-9]+\.?[a-z0-9]+\.[a-z]*\//, "/")

Answer 6

/http(?:s?):\/\/[^\/]*(\/.*)/

Answer 7

Answer 8

顶4楼此路不通