Why so difficult?

𝐀 = url.split('#')[0].split('?')[0].split('/').pop();
// Returns "file.name.ext"

𝐁 = url.split('#')[0].split('?')[0].split('/').pop().split('.')[0];
// Returns "file"; To handle multiple dots, use the last method below

The RegEx below will normally result in the same output as from the above, but it will return an empty string if the URL is significantly malformed.

𝐂 = (url.match(/^\w+:(\/+([^\/#?\s]+)){2,}(#|\?|$)/)||[])[2]||'';
// Returns an empty string for relative URLs, unlike the original approach

𝐃 = (url.match(/^\w+:(\/+([^\/#?\s]+)){2,}/)||[])[2]||'';
// Ignores trailing slashes (e.g., ".../posts/?a#b" results "posts")

𝐄 = ((url.match(/^\w+:(\/+([^\/#?\s]+)){2,}/)||[])[2]?.match(/^(.*?)(\.+[^.]*)?$/)||[])[1]||'';
// Removes the extension from the output of the 𝐃.

The 𝐀, 𝐂, 𝐃 returns file😃.name from:

• my://host/path=&%dir///file😃.name?q=1/2&3=&#a/?b//?

The 𝐃 also returns dir.name from:

• my://host/path=&%dir///dir.name/?not/these/#/ones//?

The 𝐄 is similar to the 𝐃 but returns file.name.dots from:

• my://host/path=&%dir///file.name.dots..ext?q=1/2&3=&#a/?b//?

First Edit (Sep 2023) :

• Changed processing order because RFC 3986 explicitly allows fragments to have unencoded slashes. Now it handles fine even when the query("search" in JS) includes unencoded slashes(which obviously should be encoded).
• This means it properly handles any form of URL that most browsers can recognize.
• The RegEx methods have been added which include a very basic validity test. Similar concept to the original one but use a different approach.

Second Edit (Jun 2025) :

• Methods have been added to exclude extensions. Thanks to @Dan Froberg.

Try to test your URL by entering below. The 𝐃 ignores trailing slashes as described above. Please comment if any RegEx method doesn't properly handle any valid URL.

<style>*{margin:0;padding:0;box-sizing:border-box;border:none;}input{display:block;margin:.2em auto 0;padding:.8em 1.2em;width:calc(100% - 5px);background:#f8f8f8;border-radius:1em;font-size:75%;}p{margin:.8em auto 1em;padding-left:.5em;font-size:90%;color:#0008}p::after{content:attr(title);margin-left:.5em;padding:.3em;background:#fe8;border-radius:.5em;font-weight:800;color:#000;}</style><input value="my://host/path=&%dir///file😃.name..ext?q=1/2&3=&#a/?b//?" id="input"><p>𝐀</p><p>𝐁</p><p>𝐂</p><p>𝐃</p><p>𝐄</p>
<script>
((input = document.querySelector('input'), output = document.querySelectorAll('p')) => {
  const getFilename = _=> {
    output[0].title = input.value.split('#')[0].split('?')[0].split('/').pop();
    output[1].title = input.value.split('#')[0].split('?')[0].split('/').pop().split('.')[0];
    output[2].title = (input.value.match(/^\w+:(\/+([^\/#?\s]+)){2,}(#|\?|$)/)||[])[2]||'';
    output[3].title = (input.value.match(/^\w+:(\/+([^\/#?\s]+)){2,}/)||[])[2]||'';
    output[4].title = ((input.value.match(/^\w+:(\/+([^\/#?\s]+)){2,}/)||[])[2]?.match(/^(.*?)(\.+[^.]*)?$/)||[])[1]||'';
  };
  getFilename();
  input.addEventListener('input', getFilename);
  input.select();
})();
</script>

Use the match function.

function GetFilename(url)
{
   if (url)
   {
      var m = url.toString().match(/.*\/(.+?)\./);
      if (m && m.length > 1)
      {
         return m[1];
      }
   }
   return "";
}

In addition to the existing answers, I would recommend using URL() constructor (works both in browsers and Node.js) because you can be sure your URL is valid:

const url = 'https://test.com/path/photo123.png?param1=1&param2=2#hash';

let filename = '';
try {
  filename = new URL(url).pathname.split('/').pop();
} catch (e) {
  console.error(e);
}
console.log(`filename: ${filename}`);

Similar to the others, but...I've used Tom's simple script - a single line,
then you can use the filename var anywhere:
http://www.tomhoppe.com/index.php/2008/02/grab-filename-from-window-location/

var filename = location.pathname.substr(location.pathname.lastIndexOf("/")+1);

This should work for all cases

function getFilenameFromUrl(url) {
  const pathname = new URL(url).pathname;
  const index = pathname.lastIndexOf('/');
  return pathname.substring(index + 1) // if index === -1 then index+1 will be 0
}

A regex solution which accounts for URL query and hash identifier:

function fileNameFromUrl(url) {
   var matches = url.match(/\/([^\/?#]+)[^\/]*$/);
   if (matches.length > 1) {
     return matches[1];
   }
   return null;
}

JSFiddle here.

Because cases tend to fail with custom code, I looked up to the JavaScript URL class. Alas, it chokes on relative URLs! Also, it doesn't have a property to get the file name. Not epic.

There has to be a good library out there which solves this common problem. Behold URI.js. All you need is a simple statement like the following:

let file = new URI(url).filename()

Then we can create a simple function that does null checks and removes the file extension:

function fileName(url) {
   if (url === null || typeof url === 'undefined')
      return ''
   let file = new URI(url).filename() // File name with file extension
   return file.substring(0, file.lastIndexOf('.')) // Remove the extension
}

Here's a snippet with test cases to play around with. All cases pass except drive paths.

test('Dots in file name without URL', 'dotted.file.name.png', 'dotted.file.name')
test('Dots in file name with URL', 'http://example.com/file.name.txt', 'file.name')
test('Lengthy URL with parameters', '/my/folder/filename.html#dssddsdsd?toto=33&dududu=podpodpo', 'filename')
test('URL with hash', '/my/folder/filename.html#dssddsdsd', 'filename')
test('URL with query strings', '/my/folder/filename.html?toto=33&dududu=podpodp', 'filename')
test('Hash after query string', 'http://www.myblog.com/filename.php?year=2019#06', 'filename')
 test('Query parameter with file path character', 'http://www.example.com/filename.zip?passkey=1/2', 'filename')
test('Query parameter with file path character and hash', 'http://www.example.com/filename.html?lang=en&user=Aan9u/o8ai#top', 'filename')
test('Asian characters', 'http://example.com/文件名.html', '文件名')
test('URL without file name', 'http://www.example.com', '')
test('Null', null, '')
test('Undefined', undefined, '')
test('Empty string', '', '')
test('Drive path name', 'C:/fakepath/filename.csv', 'filename')

function fileName(url) {
   if (url === null || typeof url === 'undefined')
      return ''
   let file = new URI(url).filename() // File name with file extension
   return file.substring(0, file.lastIndexOf('.')) // Remove the extension
}

function test(description, input, expected) {
   let result = fileName(input)
   let pass = 'FAIL'
   if (result === expected)
      pass = 'PASS'
   console.log(pass + ': ' + description + ': ' + input)
   console.log('  =>  "' + fileName(input) + '"')
}

<script src="https://cdn.jsdelivr.net/gh/medialize/URI.js@master/src/URI.js"></script>

Results

PASS: Dots in file name without URL: dotted.file.name.png
  =>  "dotted.file.name"
PASS: Dots in file name with URL: http://example.com/file.name.txt
  =>  "file.name"
PASS: Lengthy URL with parameters: /my/folder/filename.html#dssddsdsd?toto=33&dududu=podpodpo
  =>  "filename"
PASS: URL with hash: /my/folder/filename.html#dssddsdsd
  =>  "filename"
PASS: URL with query strings: /my/folder/filename.html?toto=33&dududu=podpodp
  =>  "filename"
PASS: Hash after query string: http://www.myblog.com/filename.php?year=2019#06
  =>  "filename"
PASS: Query parameter with file path character: http://www.example.com/filename.zip?passkey=1/2
  =>  "filename"
PASS: Query parameter with file path character and hash: http://www.example.com/filename.html?lang=en&user=Aan9u/o8ai#top
  =>  "filename"
PASS: Asian characters: http://example.com/文件名.html
  =>  "文件名"
PASS: URL without file name: http://www.example.com
  =>  ""
PASS: Null: null
  =>  ""
PASS: Undefined: undefined
  =>  ""
PASS: Empty string: 
  =>  ""
FAIL: Drive path name: C:/fakepath/filename.csv
  =>  ""

This solution is for you if you're too lazy to write custom code and don't mind using a library to do work for you. It isn't for you if you want to code golf the solution.

those will not work for lenghty url like
"/my/folder/questions.html#dssddsdsd?toto=33&dududu=podpodpo"

here I expect to get "questions.html". So a possible (slow) solution is as below

fname=function(url) 
{ return url?url.split('/').pop().split('#').shift().split('?').shift():null }

then you can test that in any case you get only the filename.

fname("/my/folder/questions.html#dssddsdsd?toto=33&dududu=podpodpo")
-->"questions.html"
fname("/my/folder/questions.html#dssddsdsd")
-->"questions.html"
fname("/my/folder/questions.html?toto=33&dududu=podpodpo")
"-->questions.html"

(and it works for null)

(I would love to see a faster or smarter solution)

This answer only works in browser environment. Not suitable for node.

function getFilename(url) {
  const filename = decodeURIComponent(new URL(url).pathname.split('/').pop());
  if (!filename) return 'index.html'; // some default filename
  return filename;
}

function filenameWithoutExtension(filename) {
  return filename.replace(/^(.+?)(?:\.[^.]*)?$/, '$1');
}

Here are two functions:

• first one get filename from url
• second one get filename without extension from a full filename

For parsing URL, new an URL object should be the best choice. Also notice that URL do not always contain a filename.

Notice: This function try to resolve filename from an URL. But it do NOT guarantee that the filename is valid and suitable for use:

• Some OS disallow certain character in filename (e.g. : in windows, \0 in most OS, ...);
• Some filename may reserved by OS (e.g. CON in windows);
• Some filename may make user unhappy to handle it (e.g. a file named "--help" in Linux)

Test it out:

function getFilename(url) {
  const filename = decodeURIComponent(new URL(url).pathname.split('/').pop());
  if (!filename) return 'index.html'; // some default filename
  return filename;
}

function test(url) {
  console.log('Filename: %o\nUrl: %o', getFilename(url), url);
}

test('http://www.example.com');
test('http://www.example.com/');
test('http://www.example.com/name.txt');
test('http://www.example.com/path/name.txt');
test('http://www.example.com/path/name.txt/realname.txt');
test('http://www.example.com/page.html#!/home');
test('http://www.example.com/page.html?lang=en&user=Aan9u/o8ai#top');
test('http://www.example.com/%E6%96%87%E4%BB%B6%E5%90%8D.txt')

I'd use the substring function combined with lastIndexOf. This will allow for filenames with periods in them e.g. given http://example.com/file.name.txt this gives file.name unlike the accepted answer that would give file.

function GetFilename(url)
{
    if (url)
    {
        return url.substring(url.lastIndexOf("/") + 1, url.lastIndexOf("."));
    }
    return "";
}

This should handle anything you throw at it (absolute URLs, relative URLs, complex AWS URLs, etc). It includes an optional default or uses a psuedorandom string if neither a filename nor a default were present.

function parseUrlFilename(url, defaultFilename = null) {
    let filename = new URL(url, "https://example.com").href.split("#").shift().split("?").shift().split("/").pop(); //No need to change "https://example.com"; it's only present to allow for processing relative URLs.
    if(!filename) {
        if(defaultFilename) {
            filename = defaultFilename;
        //No default filename provided; use a pseudorandom string.
        } else {
            filename = Math.random().toString(36).substr(2, 10);
        }
    }
    
    return filename;
}

Props to @hayatbiralem for nailing the order of the split()s.

Using jQuery with the URL plugin:

var file = jQuery.url.attr("file");
var fileNoExt = file.replace(/\.(html|htm)$/, "");
// file == "th.html", fileNoExt = "th"

For node and browsers, based on @pauls answer but solving issues with hash and more defensive:

export function getFileNameFromUrl(url) {
  const hashIndex = url.indexOf('#')
  url = hashIndex !== -1 ? url.substring(0, hashIndex) : url
  return (url.split('/').pop() || '').replace(/[\?].*$/g, '')
} 

Few cases:

describe('getFileNameFromUrl', () => {

  it('absolute, hash and no extension', () => {
    expect(getFileNameFromUrl(
      'https://foo.bar/qs/bar/js-function-to-get-filename-from-url#comment95124061_53560218'))
    .toBe('js-function-to-get-filename-from-url')
  })

  it('relative, extension and parameters', () => {
    expect(getFileNameFromUrl('../foo.png?ar=8')).toBe('foo.png')
  })

  it('file name with multiple dots, hash with slash', () => {
    expect(getFileNameFromUrl('questions/511761/js-function.min.js?bar=9.9&y=1#/src/jjj?=9.9')).toBe('js-function.min.js')
  })
})

Similarly to what @user2492653 suggested, if all you want is the name of the file like Firefox gives you, then you the split() method, which breaks the string into an array of components, then all you need to do it grab the last index.

var temp = url.split("//");
if(temp.length > 1)
 return temp[temp.length-1] //length-1 since array indexes start at 0

This would basically break C:/fakepath/test.csv into {"C:", "fakepath", "test.csv"}

my 2 cents

the LastIndexOf("/") method in itself falls down if the querystrings contain "/"

We all know they "should" be encoded as %2F but it would only take one un-escaped value to cause problems.

This version correctly handles /'s in the querystrings and has no reliance on .'s in the url

function getPageName() {
    //#### Grab the url
    var FullUrl = window.location.href;

    //#### Remove QueryStrings
    var UrlSegments = FullUrl.split("?")
    FullUrl = UrlSegments[0];

    //#### Extract the filename
    return FullUrl.substr(FullUrl.lastIndexOf("/") + 1);
}

Try this

url.substring(url.lastIndexOf('/')+1, url.length)

url? url.substring(url.lastIndexOf('/')+1, url.lastIndexOf('.')):''

• Safety is as asked for. when url is null or undefined the result is ''.
• Removes all of the path with ':', dots and any symbol including the last '/'.
• This gives the true answer 'th' as asked and not 'th.index'. That is very important of course to have it work at all.

• It allows filename to have several periods!

• Not asked, but you can also have a query string without '/' and '.'

It is a corrected answer from Abhishek Sharma so I gave him an upvote. So genious and minimal one-liner - I saw it there :)

function getFileNameWithoutExtension(url) {
  if (typeof url !== 'string') throw new Error('url must be a string');
  // Remove the QueryString
  return url.replace(/\?.*$/, '')
  // Extract the filename
  .split('/').pop()
  // Remove the extension
  .replace(/\.[^.]+$/, '');
}

This will return news from this URL http://www.myblog.com/news.php?year=2019#06.

ES6 syntax based on TypeScript

Actually, the marked answer is true but if the second if doesn't satisfy the function returns undefined, I prefer to write it like below:

const getFileNameFromUrl = (url: string): string => {
  if (url) {
    const tmp = url.split('/');
    const tmpLength = tmp.length;

    return tmpLength ? tmp[tmpLength - 1] : '';
  }

  return '';
};

For my problem, I need to have the file extension.

Simple Function (Using RegEx Pattern)

function pathInfo(s) {
    s=s.match(/(.*?\/)?(([^/]*?)(\.[^/.]+?)?)(?:[?#].*)?$/);
    return {path:s[1],file:s[2],name:s[3],ext:s[4]};
}

var sample='/folder/another/file.min.js?query=1';
var result=pathInfo(sample);
console.log(result);
/*
{
  "path": "/folder/another/",
  "file": "file.min.js",
  "name": "file.min",
  "ext": ".js"
}
*/
console.log(result.name);

Using RegEx

Let´s say you have this url:

http://127.0.0.1:3000/pages/blog/posts/1660345251.html

Using the following line of code:

var filename = location.pathname.replace(/[\D]/g, "");

Will return:

1660345251

• Notice that this number is Unix Time, which returns your local time no matter where you are in the World (this should give you really unique post names for this blog example).
• .replace(/[\D]/g, ""), replaces any non-digit character with an empty string. /[\D]/g says non-digit, and "" says empty string. More about it: here for numbers and here for letters.
• More about RegEx, here. There are lots of RegEx tools out there that can help you out to get better results in the returning value for filename.

Extra code: Unix Time to Local Time

var humanDate = new Date(0);
var timestamp = entries[index].timestamp;
humanDate.setUTCSeconds(timestamp);

humanDate is for my local time:

Fri Aug 12 2022 20:00:51 GMT-0300 (Argentina Standard Time)

Credits for this code, here.

function getFilenameFromUrlString(url) {
 const response = {
   data: {
     filename: ""
   },
   error: ""
 };
 try {
   response.data.filename = new URL(url).pathname.split("/").pop();
 } catch (e) {
   response.error = e.toString();
 }
 return response;
}

For tests check this: https://codesandbox.io/s/get-filename-from-url-string-wqthx1

if any lazy person just wants the file name without extension,here it is

const url='https://test.s3.us-west-2.amazonaws.com/d.pdf';
const filename = url.split('/').pop().split('.')[0] //187392504

Other way to convert to URL first:

function getFilenameFromUrl(url: string): string | null {
    try {
        const parsedUrl = new URL(url);
        const pathname = parsedUrl.pathname;
        const pathComponents = pathname.split('/');
        const filename = pathComponents[pathComponents.length - 1]; // Get the last component (filename)
        return filename;
    } catch (error) {
        console.error('Error parsing URL:', error);
        return null;
    }
}

// Example usage:
const url = "https://example.com/api/alerting/image/f1635b2b-7946-422a-85c1-aaf52f0e944a/1f5f648d-cc0d-4c95-b06f-eb28b770156d8436546758293928039.zip";
const filename = getFilenameFromUrl(url);

if (filename) {
    console.log('Filename extracted:', filename);
} else {
    console.log('Failed to extract filename from URL');
}

from How to get the file name from a full path using JavaScript?

var filename = fullPath.replace(/^.*[\\\/]/, '')