Question 217628 in Javascript Question posted in
A very good W3school tutorial can be found here.

Javascript regex for url detect and split

I have the below urls :

https://comanage.example.edu/sp
https://wiki.cs.example.org/sp
https://intranet.math.example.edu/sp
https://myapp.example.com/sp

For these urls, I need to define a function that can detect them as URLs and then replace the https:// and sp path from them. Basically, I just need the hostname only.For ex, like below

https://comanage.example.edu/sp      ->comanage.example.edu
https://wiki.cs.example.org/sp       ->wiki.cs.example.org
https://intranet.math.example.edu/sp ->intranet.math.example.edu
https://myapp.example.com/sp         ->myapp.example.com

For non urls, the function should detect and do not replace for them. Like below,

nonurl.example.com -> ***no replacement***

Can anyone please suggest me a solution for above,I do not have much knowledge on usage of regex.

Tags:

2

Answers


  1. 0

    The pattern ^https?:// should work here pretty easily for you. We can use that to replace http:// and https:// at the beginning of any string with an empty string

    In the pattern the ^ symbol denotes the beginning of the string. This means if http:// somehow appeared in the middle of the string, it would not match that since it must be at the beginning

    The ? marks the previous character as optional. In the pattern the s is optional to that http and https can be found.

    The / is there simply because the slashes have to be escaped.

    const urls = [
  'https://comanage.example.edu/sp',
  'https://wiki.cs.example.org/sp',
  'https://intranet.math.example.edu/sp',
  'https://myapp.example.com/sp',
  'nonurl.example.com',
];

const pattern = /^https?:///i;

urls.forEach(u => {
  console.log(u, '-->', u.replace(pattern, ''));
});
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  2. 0

    Simply use URL API

    [ 'https://comanage.example.edu/sp'
, 'https://wiki.cs.example.org/sp'
, 'https://intranet.math.example.edu/sp'
, 'https://myapp.example.com/sp' 
]
.forEach( url => console.log( new URL(url).host ) );
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