代码之家 › 专栏 › 技术社区 › Code Guy

匹配url并检查该url是否为域名中的一个

javascript

Code Guy · 技术社区 · 3 年前

我在一个数组中有一个列入黑名单的URL列表

var exclusion = ["facebook.com","instagram.com","twitter.com","youtube.com","linkedin.com","google.com","wordpress.org","pinterest.com","plus.google.com","miit.gov.cn","whatsapp.com","apple.com","goo.gl","qq.com","policies.google.com","youtu.be","microsoft.com","maps.google.com","play.google.com","wa.me","accounts.google.com","github.com","en.wikipedia.org","support.google.com"]

我会得到一个像这样的URL来测试排除列表

https://www.facebook.com
http://www.facebook.com
https://facebook.com
http://facebook.com
http://facebook.com?login=true
http://facebook.com/?login=true
instagram.com

因此

for(var i=0;i<exclusion.length;i++)
{
if("https://www.facebook.com".indexOf(exclusion[i]) == 0)
 return true;
}

是一种效率很低的技术由于该列表 “域名” 并且给定的字符串是URL

如果指定的URL在域列表中,我如何使函数返回true。

3 回复 | 直到 3 年前

Sukhjinder Singh 3 年前

你可以像下面这样匹配URL

 var exclusion = ["facebook.com","instagram.com","twitter.com","youtube.com","linkedin.com","google.com","wordpress.org","pinterest.com","plus.google.com","miit.gov.cn","whatsapp.com","apple.com","goo.gl","qq.com","policies.google.com","youtu.be","microsoft.com","maps.google.com","play.google.com","wa.me","accounts.google.com","github.com","en.wikipedia.org","support.google.com"]

  var url = "https://www.facebook.com";
  
  for(var i=0;i<exclusion.length;i++)
  {
    if(url.includes(exclusion[i])) {
     console.log('yes');
    }
    else{
    console.log("no");
    }
  }

要在匹配URL时打破循环,可以使用 break; 在 if

var exclusion = ["facebook.com","instagram.com","twitter.com","youtube.com","linkedin.com","google.com","wordpress.org","pinterest.com","plus.google.com","miit.gov.cn","whatsapp.com","apple.com","goo.gl","qq.com","policies.google.com","youtu.be","microsoft.com","maps.google.com","play.google.com","wa.me","accounts.google.com","github.com","en.wikipedia.org","support.google.com"]
    
      var url = "https://www.facebook.com";
      
      for(var i=0;i<exclusion.length;i++)
      {
        if(url.includes(exclusion[i])) {
         console.log('yes');
         break;
        }
        else{
        console.log("no");
        }
      }

Nick SamSmith1986 3 年前

一种方法是去掉URL的无关部分(最好使用URL API,但如果不可能,则使用regex),然后测试结果是否在排除数组中:

const exclusion = ["facebook.com","instagram.com","twitter.com","youtube.com","linkedin.com","google.com","wordpress.org","pinterest.com","plus.google.com","miit.gov.cn","whatsapp.com","apple.com","goo.gl","qq.com","policies.google.com","youtu.be","microsoft.com","maps.google.com","play.google.com","wa.me","accounts.google.com","github.com","en.wikipedia.org","support.google.com"]

const tests = ['https://www.facebook.com','http://www.facebook.com','https://facebook.com','http://facebook.com','http://facebook.com?login=true','http://facebook.com/?login=true','instagram.com']

var urlRegex = /^https?:\/\/(?:www\.)?([^/?]+).*$/

const blacklistedURL = (url) => exclusion.includes(url.replace(urlRegex, '$1'))

tests.forEach(url => {
  if (blacklistedURL(url)) {
    console.log(`${url} is blacklisted!`)
  }
})

笔记出于演示的目的,我在代码中使用了一个简单的URL匹配正则表达式。有许多更好的正则表达式来匹配URL,您应该使用其中之一。

KARTHICK RAVI 3 年前

为了进行检查,我们可以使用includes。我已经通过数组附加了代码映像映射并进行了检查。希望有帮助