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使用正则表达式在字符串第n次出现后捕获

regex node.js javascript

1

S16 · 技术社区 · 4 年前

我的测试字符串:

/custom-heads/food-drinks/51374-easter-bunny-cake

我试图捕捉字符串中的数字。该字符串中的常量是数字,数字前面总是有3 / 然后是一个 - .

我是一个regex noob,正在努力解决这个问题。我拼凑起来 (\/)(.*?)(-) 然后我想我可以通过编程得到最后一个,但我真的很想更好地理解正则表达式,如果有人能给我看正则表达式,让我得到最后一个出现在 / 和 - .

3 回复 | 直到 4 年前

1

Maielo 4 年前

如果可能的话,不要使用正则表达式,我建议你阅读- https://blog.codinghorror.com/regular-expressions-now-you-have-two-problems/ 博客帖子

对于你的问题,使用劈裂更容易、更快、更防弹

const articleName = "/custom-heads/food-drinks/51374-easter-bunny-cake".split("/")[3]
// '51374-easter-bunny-cake'

const articleId = articleName.split("-")[0]

// '51374'

希望有帮助

2

0

anubhava 4 年前

您可以将此正则表达式与捕获组一起使用:

^(?:[^\/]*\/){3}([^-]+)

或者在现代浏览器中,可以使用lookback断言:

/(?<=^(?:[^\/]*\/){3})[^-]+/

RegEx Demo 1

RegEx Demo 2

正则表达式代码:

^ :开始
(?:[^\/]*\/){3} :匹配0个或多个非- / 字符后跟 / .重复这组动作3次
([^-]+) :匹配1+个非连字符

代码:

const s = `/custom-heads/food-drinks/51374-easter-bunny-cake`;

const re = /^(?:[^\/]*\/){3}([^-]+)/;

console.log (s.match(re)[1]);

3

0

Ryszard Czech 4 年前

使用

const str = `/custom-heads/food-drinks/51374-easter-bunny-cake`
const p = /(?:\/[^\/]*){2}\/(\d+)-/
console.log(str.match(p)?.[1])

看见 regex proof .

解释

Non-capturing group (?:\/[^\/]*){2}
{2} matches the previous token exactly 2 times
\/ matches the character / with index 4710 (2F16 or 578) literally (case sensitive)
  Match a single character not present in the list below [^\/]
  * matches the previous token between zero and unlimited times, as many times as possible, giving back as needed (greedy)
  \/ matches the character / with index 4710 (2F16 or 578) literally (case sensitive)
\/ matches the character / with index 4710 (2F16 or 578) literally (case sensitive)
1st Capturing Group (\d+)
  \d matches a digit (equivalent to [0-9])
  + matches the previous token between one and unlimited times, as many times as possible, giving back as needed (greedy)
- matches the character - with index 4510 (2D16 or 558) literally (case sensitive)