使用JS正则表达式获取没有特定后缀的字符串段

Question

我有个网址，可以是

https://arxiv.org/abs/2004.10934

或带有后缀v0-9。

https://arxiv.org/abs/2004.10934v3

我想用regex过滤掉v0-9部分。

也就是说，无论我们给哪种类型的url，都会得到https://arxiv.org/abs/2004.10934'。

下面是我现在所拥有的，它很有效，但看起来很烦躁.

let url = 'https://arxiv.org/abs/2004.10934v3'
let regURL = /(.*(?<!v[0-9])(?<!v))/g;
let url_f = regURL.exec(url)[0];

有更好的正则表达式吗？

Answer 1

使用

/^(.*?)(v\d+)?$/

见证明。

解释

NODE                     EXPLANATION
--------------------------------------------------------------------------------
  ^                        the beginning of the string
--------------------------------------------------------------------------------
  (                        group and capture to \1:
--------------------------------------------------------------------------------
    .*?                      any character except \n (0 or more times
                             (matching the least amount possible))
--------------------------------------------------------------------------------
  )                        end of \1
--------------------------------------------------------------------------------
  (                        group and capture to \2 (optional
                           (matching the most amount possible)):
--------------------------------------------------------------------------------
    v                        'v'
--------------------------------------------------------------------------------
    \d+                      digits (0-9) (1 or more times (matching
                             the most amount possible))
--------------------------------------------------------------------------------
  )?                       end of \2 (NOTE: because you are using a
                           quantifier on this capture, only the LAST
                           repetition of the captured pattern will be
                           stored in \2)
--------------------------------------------------------------------------------
  $                        before an optional \n, and the end of the
                           string

JavaScript：

​

let url = 'https://arxiv.org/abs/2004.10934v3'
let regURL = /^(.*?)(v\d+)?$/;
let [_, url_f, version] = regURL.exec(url);
console.log(url_f);

​

Answer 2

如果用空字符串替换v0-9，那么它将被过滤掉：

url.replace(/v[0-9]$/, '')