js function to get filename from url

Question

I have a url like http://www.example.com/blah/th.html

I need a javascript function to give me the 'th' value from that.

All my urls have the same format (2 letter filenames, with .html extension).

I want it to be a safe function, so if someone passes in an empty url it doesn't break.

I know how to check for length, but I should be checking for null to right?

Answer 1

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');
}

Answer 2

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

Answer 3

Why so difficult?

= url.split('#')[0].split('?')[0].split('/').pop();

RegEx below would result same as the above's normally, but will return empty string if the URL was significantly malformed.

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

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

---

All three of them would return file😃.name from:

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

The third one would also return dir.name from:

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

---

First Edit (Sep 2023) :

• Changed processing order because RFC 3986 explicitly allows fragments to having unencoded slashes. Now it would handles fine even the query("search" in JS) includes unencoded slashes(which obviously had to be encoded).
• This means it might properly handles any form of URLs that most browsers could recognize.
• Using RegEx method was added which includes very basic validity test. Similar concept to the original one but using different approach.

---

Try to test your URL by enter below. The third one would ignores trailing slash as described the above code. Please comment if the RegEx ones wouldn't properly handle any valid URL

<style>*{margin:0;padding:0;box-sizing:border-box;border:none;}label{font-size:90%}input{display:block;margin:.2em auto 0;padding:.8em 1.2em;width:calc(100% - 5px);background:#f8f8f8;border-radius:1em;}p{margin-top:1.4em;padding-left:.5em;font-size:95%;color:#0008}p::after{content:attr(title);margin-left:.5em;padding:.4em .2em;background:#fe8;border-radius:.5em;font-weight:800;color:#000;}</style>
<label for="input">Enter your URL here:</label>
<input value="my://host/path=&%dir///file😃.name?q=1/2&3=&#a/?b//?" id="input">
<p>A</p>
<p>B</p>
<p>C</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.match(/^\w+:(\/+([^\/#?\s]+)){2,}(#|\?|$)/)||[])[2]||'';
    output[2].title = (input.value.match(/^\w+:(\/+([^\/#?\s]+)){2,}/)||[])[2]||'';
  };
  getFilename();
  input.addEventListener('input', getFilename);
  input.select();
})();
</script>

Answer 4

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

Answer 5

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.

Answer 6

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
}

Answer 7

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);

Answer 8

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.

Answer 9

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}`);

Answer 10

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.

Answer 11

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.

Answer 12

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')

Answer 13

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')
  })
})

Answer 14

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.

Answer 15

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 :)

Answer 16

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

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

Answer 17

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.

Answer 18

Try this

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

Answer 19

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);
}

Answer 20

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)

Answer 21

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);

Answer 22

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"}

Answer 23

var filename = url.split('/').pop()

Answer 24

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 "";
}

Answer 25

Using jQuery with the URL plugin:

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

Answer 26

Use the match function.

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