regex - Remove everything after domain and http in url javascript - Stack Overflow

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I have the full url in a javascript variable and I want to strip it down to the bare url.

For example:

.php,

strip down to:

www.google, youtube

How would I do this in javascript?

Thanks

This is not similar to the other links as I am doing this within a chrome extension, therefore the only way to get the url is using the chrome extension api which only provides the full url. As such I need to strip the full url

I have the full url in a javascript variable and I want to strip it down to the bare url.

For example:

https://www.google./hello/hi.php, http://youtube.

strip down to:

www.google., youtube.

How would I do this in javascript?

Thanks

This is not similar to the other links as I am doing this within a chrome extension, therefore the only way to get the url is using the chrome extension api which only provides the full url. As such I need to strip the full url

• javascript
• regex
• string
• strip

Share Improve this question Follow edited Aug 11, 2015 at 12:41 kabeersvohra asked Aug 11, 2015 at 12:32 kabeersvohrakabeersvohra 1,06922 gold badges1414 silver badges3131 bronze badges 0

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3 Answers 3

Sorted by: Reset to default 5

You can try this:

\/\/([^\/,\s]+\.[^\/,\s]+?)(?=\/|,|\s|$|\?|#)

Regex live here.

---

Live JavaScript sample:

var regex = /\/\/([^\/,\s]+\.[^\/,\s]+?)(?=\/|,|\s|$|\?|#)/g;

var input = "https://www.google./hello/hi.php, http://youtube.,"
          + "http://test#jump or http://google.?q=test";

while (match = regex.exec(input)) {
    document.write(match[1] + "<br/>");
};

---

Hope it helps

Use the global window.location.hostname variable, and it will give you this information.

While parsing with regular expressions like the other answers suggest is possible, a better approach would be to use the URL() object/API (docs).

const a = "https://www.google./hello/hi.php";
const hostname = new URL(a).hostname; // "www.google."

Apart from being arguably "cleaner", another advantage is that if the underlying parsing logic is incorrect, it is a (potentially security) bug in the browser, not your code, and will be likely fixed without your effort.

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