blocks|key|1967890|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1967891|var+url+=+"https://site.com";
var+urlNoProtocol+=+url.replace(/%5Ehttps?\:\/\//i,+"");|code-block|syntax|javascript|1967892|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Try with this:

<pre><code>var url = "https://site.com";
var urlNoProtocol = url.replace(/^https?\:\/\//i, "");
</code></pre>

blocks|key|1218664|text|您可以像这样使用URL对象：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1218665|const+urlWithoutProtocol+=+new+URL(url).host;|style|CODE|1218666|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.mozilla.org/en-US/docs/Web/API/URL/URL^0|8|3|0|0|0|19|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@$A|Q|B|R|1|S]]|C|$]]|$1|D|3|E|5|6|7|T|8|@$A|U|B|V|F|G]]|9|@]|C|$]]|$1|H|3|-4|5|6|7|W|8|@]|9|@]|C|$]]]|I|$J|$5|K|L|M|C|$N|O]]]]

You can use the <a href="https://developer.mozilla.org/en-US/docs/Web/API/URL/URL" rel="noreferrer">URL</a> object like this: 

<code>const urlWithoutProtocol = new URL(url).host;</code>

blocks|key|1967869|text|var+txt="https://site.com";
txt=/%5Ehttp(s)?:\/\/(.%2B)$/i.exec(txt);
txt=txt[2];|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1967870|要解析没有http/https的链接，请使用以下命令：|unstyled|1967871|var+txt="https://site.com";
txt=/%5E(http(s)?:\/\/)?(.%2B)$/i.exec(txt);
txt=txt[3];|1967872|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>var txt="https://site.com";
txt=/^http(s)?:\/\/(.+)$/i.exec(txt);
txt=txt[2];
</code></pre>

for parsing links without http/https use this:

<pre><code>var txt="https://site.com";
txt=/^(http(s)?:\/\/)?(.+)$/i.exec(txt);
txt=txt[3];
</code></pre>

blocks|key|3403324|text|您可以使用URL()+constructor。它将解析您的url字符串，并且将有一个没有协议的条目。因此，使用正则表达式可以减少头痛：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|3403325|let+u+=+new+URL('https://www.facebook.com/companypage/');
URL+{
++++hash:+""
++++host:+"www.facebook.com"
++++hostname:+"www.facebook.com"
++++href:+"https://www.facebook.com/companypage/"
++++origin:+"https://www.facebook.com"
++++password:+""
++++pathname:+"/companypage/"
++++port:+""
++++protocol:+"https:"
++++search:+""
++++searchParams:+URLSearchParams+{}
++++username:+""
}
u.host+//+www.facebook.com
u.hostname+//+www.facebook.com|code-block|syntax|javascript|3403326|尽管URL()去掉了一个协议，但它给您留下了www+part。在我的例子中，我也想摆脱那个子域部分，所以无论如何都要用到.replace()。|style|CODE|3403327|u.host.replace(/%5Ewww./,+'')+//+www.facebook.com+=>+facebook.com|3403328|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.mozilla.org/en-US/docs/Web/API/URL/URL^0|5|H|0|0|0|2|5|M|3|1O|A|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|11|8|@$A|12|B|13|K|L]|$A|14|B|15|K|L]|$A|16|B|17|K|L]]|9|@]|C|$]]|$1|M|3|N|5|F|7|18|8|@]|9|@]|C|$G|H]]|$1|O|3|-4|5|6|7|19|8|@]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]]]

You may use <a href="https://developer.mozilla.org/en-US/docs/Web/API/URL/URL" rel="nofollow noreferrer">URL() constructor</a>. It will parse your url string and there will be an entry w/o protocol. So less headache with regexps:

<pre><code>let u = new URL('https://www.facebook.com/companypage/');
URL {
 hash: ""
 host: "www.facebook.com"
 hostname: "www.facebook.com"
 href: "https://www.facebook.com/companypage/"
 origin: "https://www.facebook.com"
 password: ""
 pathname: "/companypage/"
 port: ""
 protocol: "https:"
 search: ""
 searchParams: URLSearchParams {}
 username: ""
}
u.host // www.facebook.com
u.hostname // www.facebook.com
</code></pre>

Although <code>URL()</code> drops out a protocol, it leaves you with <code>www</code> part. In my case I wanted to get rid of that subdomain part as well, so had to use to <code>.replace()</code> anyway.

<pre><code>u.host.replace(/^www./, '') // www.facebook.com =&gt; facebook.com
</code></pre>

blocks|key|3403095|text|var+str+=+"https://site.com";

str+=+str.substr(+str.indexOf(':')+%2B+3+);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3403096|您也可以使用.slice()或.substring()来代替.substr()。在这种情况下，它们都会产生相同的结果。|unstyled|offset|length|style|CODE|3403097|str+=+str.slice(+str.indexOf(':')+%2B+3+);

str+=+str.substring(+str.indexOf(':')+%2B+3+);|3403098|3403099|EDIT:在另一个答案下的评论中，问题的要求似乎发生了变化。|BOLD|3403100|如果字符串中可能没有http://，则执行以下操作：|3403101|var+str+=+"site.com";

var+index+=+str.indexOf('://');
if(+index+>+-1+)
+++str+=+str.substr(+index+%2B+3+);|3403102|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/String/slice|1|https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/String/substring|2|https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/String/substr^0|0|6|8|F|C|U|9|6|8|0|F|C|1|U|9|2|0|0|0|0|5|0|A|7|0|0^^$0|@$1|2|3|4|5|6|7|16|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|17|8|@$G|18|H|19|I|J]|$G|1A|H|1B|I|J]|$G|1C|H|1D|I|J]]|9|@$G|1E|H|1F|1|1G]|$G|1H|H|1I|1|1J]|$G|1K|H|1L|1|1M]]|A|$]]|$1|K|3|L|5|6|7|1N|8|@]|9|@]|A|$B|C]]|$1|M|3|-4|5|F|7|1O|8|@]|9|@]|A|$]]|$1|N|3|O|5|F|7|1P|8|@$G|1Q|H|1R|I|P]]|9|@]|A|$]]|$1|Q|3|R|5|F|7|1S|8|@$G|1T|H|1U|I|J]]|9|@]|A|$]]|$1|S|3|T|5|6|7|1V|8|@]|9|@]|A|$B|C]]|$1|U|3|-4|5|F|7|1W|8|@]|9|@]|A|$]]]|V|$W|$5|X|Y|Z|A|$10|11]]|12|$5|X|Y|Z|A|$10|13]]|14|$5|X|Y|Z|A|$10|15]]]]

<pre><code>var str = "https://site.com";

str = str.substr( str.indexOf(':') + 3 );
</code></pre>

Instead of <a href="https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/String/substr" rel="nofollow"><code>.substr()</code></a>, you could also use <a href="https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/String/slice" rel="nofollow"><code>.slice()</code></a> or <a href="https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/String/substring" rel="nofollow"><code>.substring()</code></a>. They'll all produce the same result in this situation.

<pre><code>str = str.slice( str.indexOf(':') + 3 );

str = str.substring( str.indexOf(':') + 3 );
</code></pre>

<hr>

EDIT: It appears as though the requirements of the question have changed in a comment under another answer.

If there possibly isn't a <code>http://</code> in the string, then do this:

<pre><code>var str = "site.com";

var index = str.indexOf('://');
if( index &gt; -1 )
 str = str.substr( index + 3 );
</code></pre>

blocks|key|4814412|text|从URL中剥离协议：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4814413|var+url+=+"https://site.com";
var+urlNoProto+=+url.split('/').slice(2).join('/');|code-block|syntax|javascript|4814414|适用于任何协议，ftp，http，gopher，nntp，telnet，wais，file，prospero+.RFC+1738中指定的所有类型，但其中没有//的类型除外(mailto、news)。|4814415|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Strip the protocol from a URL:

<pre><code>var url = "https://site.com";
var urlNoProto = url.split('/').slice(2).join('/');
</code></pre>

Works with any protocol, ftp, http, gopher, nntp, telnet, wais, file, prospero ... all those specified in RFC 1738 with the exception of those without a // in them (mailto, news).

blocks|key|4814535|text|请注意，在实际网页中，继承协议+//是一种常见的实践https://paulirish.com/2010/the-protocol-relative-url。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|CODE|entityRanges|data|4814536|所以我建议regexp也涵盖这个案例：|4814537|/%5E\/\/%7C%5Ehttps?:\/\//|4814538|(可以对其进行优化)|4814539|entityMap|0|LINK|mutability|MUTABLE|url|https://paulirish.com/2010/the-protocol-relative-url^0|B|4|G|2|Q|1G|0|0|0|0|K|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|D]]|E|@$9|Z|A|10|1|11]]|F|$]]|$1|G|3|H|5|6|7|12|8|@]|E|@]|F|$]]|$1|I|3|J|5|6|7|13|8|@$9|14|A|15|B|D]]|E|@]|F|$]]|$1|K|3|L|5|6|7|16|8|@]|E|@]|F|$]]|$1|M|3|-4|5|6|7|17|8|@]|E|@]|F|$]]]|N|$O|$5|P|Q|R|F|$S|T]]]]

Please note that in real web pages inherited protocol <code>//</code> is a common practice <a href="https://paulirish.com/2010/the-protocol-relative-url" rel="nofollow noreferrer">https://paulirish.com/2010/the-protocol-relative-url</a>. 

So I suggest regexp covering this case as well: 

<code>/^\/\/|^https?:\/\//</code> 

(you can optimize it)

blocks|key|3403343|text|这个答案扩展了上面的一些答案，http://，https://，或+//，这也很常见。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|BOLD|entityRanges|data|3403344|感谢上面的答案让我想到了这一点！|3403345|const+urls+=+[+"http://example.com",+"https://example.com",+"//example.com"+]

//+the+regex+below+states:+replace+`//`+or+replace+`//`+and+the+'stuff'
const+resolveHostNames+=+urls.map(url+=>+url.replace(/\/\/%7C.%2B\/\//,+''))

console.log(resolveHostNames);|code-block|syntax|javascript|3403346|这是一个link+to+a+codepen。|3403347|entityMap|0|LINK|mutability|MUTABLE|url|https://codepen.io/yowainwright/pen/eydvPv/^0|F|7|N|8|Y|2|W|1|0|0|0|4|H|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@$9|Y|A|Z|B|C]|$9|10|A|11|B|C]|$9|12|A|13|B|C]|$9|14|A|15|B|D]]|E|@]|F|$]]|$1|G|3|H|5|6|7|16|8|@]|E|@]|F|$]]|$1|I|3|J|5|K|7|17|8|@]|E|@]|F|$L|M]]|$1|N|3|O|5|6|7|18|8|@]|E|@$9|19|A|1A|1|1B]]|F|$]]|$1|P|3|-4|5|6|7|1C|8|@]|E|@]|F|$]]]|Q|$R|$5|S|T|U|F|$V|W]]]]

This answer extends some answers above, <code>http://</code>, <code>https://</code>, or <code>//</code> which is also common. 

Thanks for answers above that led me to this!

<pre><code>const urls = [ "http://example.com", "https://example.com", "//example.com" ]

// the regex below states: replace `//` or replace `//` and the 'stuff'
const resolveHostNames = urls.map(url =&gt; url.replace(/\/\/|.+\/\//, ''))

console.log(resolveHostNames);
</code></pre>

Here's a <a href="https://codepen.io/yowainwright/pen/eydvPv/" rel="nofollow noreferrer">link to a codepen</a>.

blocks|key|1218660|text|另一种有效的解决方案，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1218661|url.replace(/(%5E(\w%2B:)?\/\//,+'')|offset|length|style|CODE|1218662|entityMap^0|0|0|W|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

Another efficient solution, 

<code>url.replace(/(^(\w+:)?\/\//, '')</code>

blocks|key|1218554|text|假设除了协议之外没有双斜杠，您可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1218555|+var+url+=+"https://example.com";
+var+noProtocol+=+url.split('//')[1];|code-block|syntax|javascript|1218556|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Assuming there are no double slashes other than the protocol, you could do:

<pre><code> var url = "https://example.com";
 var noProtocol = url.split('//')[1];
</code></pre>

blocks|key|4814483|text|您可以使用DOM的HTMLHyperlinkElementUtils：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4814484|function+removeProtocol(url)+{
++const+a+=+document.createElement('a');
++a.href+=+url;
++//+`url`+may+be+relative,+but+`a.href`+will+be+absolute.
++return+a.href.replace(a.protocol+%2B+'//',+'');
}

removeProtocol('https://example.com/https://foo');
//+'example.com/https://foo'

removeProtocol('wrong://bad_example/u');
//+'bad_example/u'|code-block|syntax|javascript|4814485|来自HTMLHyperlinkElementUtils+on+MDN|offset|length|4814486|a.hostname，example.com|style|CODE|4814487|a.host，example.com:3000|4814488|a.pathname，/foo/bar.html|4814489|a.search，?a=1&b=2|4814490|a.hash，#goo|4814491|a.username、a.password、a.port等。|4814492|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.mozilla.org/en-US/docs/Web/API/HTMLHyperlinkElementUtils^0|0|0|2|W|0|0|0|A|0|0|6|0|0|A|0|0|8|0|0|6|0|0|A|B|A|M|6|0^^$0|@$1|2|3|4|5|6|7|16|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|18|8|@]|9|@$I|19|J|1A|1|1B]]|A|$]]|$1|K|3|L|5|6|7|1C|8|@$I|1D|J|1E|M|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|1F|8|@$I|1G|J|1H|M|N]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1I|8|@$I|1J|J|1K|M|N]]|9|@]|A|$]]|$1|S|3|T|5|6|7|1L|8|@$I|1M|J|1N|M|N]]|9|@]|A|$]]|$1|U|3|V|5|6|7|1O|8|@$I|1P|J|1Q|M|N]]|9|@]|A|$]]|$1|W|3|X|5|6|7|1R|8|@$I|1S|J|1T|M|N]|$I|1U|J|1V|M|N]|$I|1W|J|1X|M|N]]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|1Y|8|@]|9|@]|A|$]]]|Z|$10|$5|11|12|13|A|$14|15]]]]

You may use HTMLHyperlinkElementUtils of DOM:


<pre><code>function removeProtocol(url) {
 const a = document.createElement('a');
 a.href = url;
 // `url` may be relative, but `a.href` will be absolute.
 return a.href.replace(a.protocol + '//', '');
}

removeProtocol('https://example.com/https://foo');
// 'example.com/https://foo'

removeProtocol('wrong://bad_example/u');
// 'bad_example/u'
</code></pre>

From <a href="https://developer.mozilla.org/en-US/docs/Web/API/HTMLHyperlinkElementUtils" rel="nofollow">HTMLHyperlinkElementUtils on MDN</a>:

<code>a.hostname</code>, example.com 
<code>a.host</code>, example.com:3000 
<code>a.pathname</code>, /foo/bar.html 
<code>a.search</code>, ?a=1&amp;b=2 
<code>a.hash</code>, #goo 
<code>a.username</code>, <code>a.password</code>, <code>a.port</code>, etc.

blocks|key|1963358|text|如果有一个方便的原生URL接口，只需2行代码就可以完成这项工作，那么使用正则表达式可能就有点过分了：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1963359|​|1963360|let+url+=+"https://stackoverflow.com/questions/3999764/taking-off-the-http-or-https-off-a-javascript-string";
let+a+=+new+URL(url);
let+withoutProtocol+=+a.host%2Ba.pathname;
console.log(`Without+protocol:+${withoutProtocol}`);
console.log(`With+protocol:+${url}`);|code-block|syntax|javascript|1963361|1963362|URL+API+Support+in+browsers|1963363|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.mozilla.org/en-US/docs/Web/API/URL|1|https://caniuse.com/url^0|A|3|A|3|0|0|0|0|0|0|R|1|0^^$0|@$1|2|3|4|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@$9|12|A|13|1|14]]|E|$]]|$1|F|3|G|5|6|7|15|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|16|8|@]|D|@]|E|$K|L]]|$1|M|3|G|5|6|7|17|8|@]|D|@]|E|$]]|$1|N|3|O|5|6|7|18|8|@]|D|@$9|19|A|1A|1|1B]]|E|$]]|$1|P|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|Q|$R|$5|S|T|U|E|$V|W]]|X|$5|S|T|U|E|$V|Y]]]]

Using regex might be an overkill when there's a handy native <a href="https://developer.mozilla.org/en-US/docs/Web/API/URL" rel="nofollow noreferrer"><code>URL</code></a> interface that does the job for you in 2 lines:
<div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>let url = "https://stackoverflow.com/questions/3999764/taking-off-the-http-or-https-off-a-javascript-string";
let a = new URL(url);
let withoutProtocol = a.host+a.pathname;
console.log(`Without protocol: ${withoutProtocol}`);
console.log(`With protocol: ${url}`);</code></pre>
</div>
</div>

<a href="https://caniuse.com/url" rel="nofollow noreferrer">URL API Support in browsers</a>

blocks|key|1963303|text|Javascript对split函数的使用也破坏了解决方案。太棒了！|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1963304|var+url+=+"https://example.com";

url+=+url.split("://")[1];++++//+for+https+use+url..split("://")[0];
console.log(url);|code-block|syntax|javascript|1963305|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Javascript use of split function also dervies the solution.
Awesome !!!

<pre><code>var url = "https://example.com";

url = url.split("://")[1]; // for https use url..split("://")[0];
console.log(url);
</code></pre>

I have the following strings

<pre><code>http://example.com
https://example.com
http://www.example.com
</code></pre>

how do i get rid of the <code>http://</code> or <code>https://</code>?

taking off the http or https off a javascript string

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有以下字符串http://example.comhttps://example.comhttp://www.example.com如何摆脱http://或https://

问从javascript字符串中去掉http或https
EN

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问从javascript字符串中去掉http或httpsEN

回答 13

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问从javascript字符串中去掉http或https
EN