如何拆分任何url并只获取源

Question

我想拆分https://abcnews.go.com/这种类型的URL。

想要获得abcnews

我用过：

.split('//')[1]
.split('www.')[1]
.split('.com')[0]
.split('.org')[0]
.split('.edu')[0]
.split('.info')[0]

我得到错误不能分开。

知道怎么做吗？

Answer 1

这看起来很安全：

(?:https?://(?:www.)?|www.)([^./]+)

见证明。

解释

--------------------------------------------------------------------------------
(?:                      group, but do not capture:
--------------------------------------------------------------------------------
http                     'http'
--------------------------------------------------------------------------------
s?                       's' (optional (matching the most amount
possible))
--------------------------------------------------------------------------------
:                        ':'
--------------------------------------------------------------------------------
/                       '/'
--------------------------------------------------------------------------------
/                       '/'
--------------------------------------------------------------------------------
(?:                      group, but do not capture (optional
(matching the most amount possible)):
--------------------------------------------------------------------------------
www                      'www'
--------------------------------------------------------------------------------
.                       '.'
--------------------------------------------------------------------------------
)?                       end of grouping
--------------------------------------------------------------------------------
|                        OR
--------------------------------------------------------------------------------
www                      'www'
--------------------------------------------------------------------------------
.                       '.'
--------------------------------------------------------------------------------
)                        end of grouping
--------------------------------------------------------------------------------
(                        group and capture to 1:
--------------------------------------------------------------------------------
[^./]+                  any character except: '.', '/' (1 or
more times (matching the most amount
possible))
--------------------------------------------------------------------------------
)                        end of 1

JavaScript代码:

const url = "https://abcnews.go.com/"
const regex = /(?:https?://(?:www.)?|www.)([^./]+)/
const match = url.match(regex)
if (match) {
console.log(match[1])
}

Answer 2

您接受Regex的回复吗？

const url='https://abcnews.go.com/';
let subdomain = url.match(/(w*)./)[1];
console.log(subdomain)
// subdomain = "abcnews"

这样，你就可以在第一个点之前得到单词。

---

我认为您在以下时间后会出现错误：.split('//')[1].split('www.')[1]因为你没有；www"；，则数组的长度为1(仅array[0])。

const rgx=/(?:www.)?(w*)./;
const abc='https://abcnews.go.com/';
const www='https://www.google.com/';
console.log(abc.match(rgx)[1]);
// "abcnews"
console.log(www.match(rgx)[1]);
// "google"