使用正则表达式从字符串中提取URL [英] Extracting for URL from string using regex

问题描述

我想提取字符串中的第一个有效URL，该字符串可以在字符和空格之间的任意位置

I want to extract the first valid URL in a string, which can be anywhere between characters and whitespace

我尝试了以下方法

...
urlRegex: /^(http[s]?:\/\/.*?\/[a-zA-Z-_]+.*)$/,

...
var input = event.target.value // <--- some string;
var url   = input.match(this.urlRegex);

问题是 url 在找到url时返回整个字符串，而不是仅返回与正则表达式匹配的字符串的一部分

Problem is url returns the whole string when it finds a url, instead of returning just the part of the string matching the regex

示例字符串

https://medium.com/aspen-ideas/there-s-no-blueprint-26f6a2fbb99c随机材料sd

返回

["https://medium.com/aspen-ideas/there-s-no-blueprint-26f6a2fbb99c random stuff sd", "https://medium.com/aspen-ideas/there-s-no-blueprint-26f6a2fbb99c random stuff sd", index: 0, input: "https://medium.com/aspen-ideas/there-s-no-blueprint-26f6a2fbb99c random stuff sd"]

如何实现?

推荐答案

您的 regex 不正确.

Your regex is incorrect.

用于提取URl的正则表达式: /(https?:\/\/[^] *)/

查看此小提琴.

这是代码段.

var urlRegex = /(https?:\/\/[^ ]*)/;

var input = "https://medium.com/aspen-ideas/there-s-no-blueprint-26f6a2fbb99c random stuff sd";
var url = input.match(urlRegex)[1];
alert(url);

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