Get protocol, domain, and port from URL

I need to extract the full protocol, domain, and port from a given URL. For example:

https://localhost:8181/ContactUs-1.0/contact?lang=it&report_type=consumer
>>>
https://localhost:8181

---

first get the current address

var url = window.location.href

Then just parse that string

var arr = url.split("/");

your url is:

var result = arr[0] + "//" + arr[2]

Hope this helps

---

var full = location.protocol+'//'+location.hostname+(location.port ? ':'+location.port: '');

---

None of these answers seem to completely address the question, which calls for an arbitrary url, not specifically the url of the current page.

Method 1: Use the URL API (caveat: no IE11 support)

You can use the URL API (not supported by IE11, but available everywhere else).

This also makes it easy to access search params. Another bonus: it can be used in a Web Worker since it doesn't depend on the DOM.

const url = new URL('http://example.com:12345/blog/foo/bar?startIndex=1&pageSize=10');

Method 2 (old way): Use the browser's built-in parser in the DOM

Use this if you need this to work on older browsers as well.

//  Create an anchor element (note: no need to append this element to the document)
const url = document.createElement('a');
//  Set href to any path
url.setAttribute('href', 'http://example.com:12345/blog/foo/bar?startIndex=1&pageSize=10');

That's it!

The browser's built-in parser has already done its job. Now you can just grab the parts you need (note that this works for both methods above):

//  Get any piece of the url you're interested in
url.hostname;  //  'example.com'
url.port;      //  12345
url.search;    //  '?startIndex=1&pageSize=10'
url.pathname;  //  '/blog/foo/bar'
url.protocol;  //  'http:'

Bonus: Search params

Chances are you'll probably want to break apart the search url params as well, since '?startIndex=1&pageSize=10' isn't too useable on its own.

If you used Method 1 (URL API) above, you simply use the searchParams getters:

url.searchParams.get('searchIndex');  // '1'

Or to get all parameters:

Array.from(url.searchParams).reduce((accum, [key, val]) => {
  accum[key] = val;
  return accum;
}, {});
// -> { startIndex: '1', pageSize: '10' }

If you used Method 2 (the old way), you can use something like this:

// Simple object output (note: does NOT preserve duplicate keys).
var parms = url.search.substr(1); // remove '?' prefix
params.split('&').reduce((accum, keyval) => {
  const [key, val] = keyval.split('=');
  accum[key] = val;
  return accum;
}, {});
// -> { startIndex: '1', pageSize: '10' }

链接地址: http://www.djcxy.com/p/3848.html

上一篇: 请求地址在JavaScript中

下一篇: 从URL获取协议，域和端口