91

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?

Blankman
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    Possible duplicate of [How to get the file name from a full path using JavaScript?](https://stackoverflow.com/questions/423376/how-to-get-the-file-name-from-a-full-path-using-javascript) – Liam Aug 02 '17 at 13:45
  • For something as specific as you want, you may use RegEx: https://stackoverflow.com/a/73341035/7389293 – carloswm85 Aug 13 '22 at 01:36

24 Answers24

217
var filename = url.split('/').pop()
Baby Groot
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user2492653
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    By far the simplest and best answer. – Gregory Bolkenstijn May 01 '15 at 15:16
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    what if there are query parameters behind? – Yangshun Tay Jan 08 '16 at 07:33
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    This also doesn't handle URL fragment IDs (i.e. # IDs) – Benjineer Apr 03 '16 at 03:02
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    How about this: url.split('#').shift().split('?').shift().split('/').pop() – hayatbiralem May 03 '16 at 19:07
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    @hayatbiralem your answer is the correct one : you have to start by cleaning # and ? and afterwards get split on / or you will get the / in querystring and anchors – systho Jan 30 '18 at 09:20
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    it is almost correct but won't work with `http://domain/THE_PATH?firstParam=BLABLA/BLEBLE".split('/').pop()` – Victor Nov 30 '18 at 15:16
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    Not correct at all despite all upvotes! Miss a lot of points.. does not check if url is undefined ... does not split on '.' ... does not remove eventual query string ... –  Jun 16 '19 at 20:25
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    Even assuming no query string, path, fragment, whatever, doesn't this produce 'th.html' and not 'th', which was what the requester asked for? – Booboo Jun 16 '19 at 22:38
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    How about this? `new URL(url).pathname.split('/').pop();` – Sangbok Lee Dec 30 '19 at 07:59
  • @Booboo You can add `.split('.')[0]` at the end. So the full line would be `var filename = url.split('/').pop().split('#')[0].split('?')[0].split('.')[0];` – TommyZG Mar 25 '22 at 23:12
  • @SangbokLee i guess what you are suggesting is what I implemented here in one of this post's answers.. you may see it here => https://stackoverflow.com/a/53560218/903998 – Victor Jul 21 '22 at 14:48
  • this solution actually doesn't strip query parameters nor anchors from the url https://i.ibb.co/c3SMNmS/image.png and actually may have some NOT correct results given some url without a path https://i.ibb.co/jJQ4CnC/image.png. It is a very simple solution that works for some cases. – Victor Sep 24 '22 at 15:04
  • url.split('/').at(-1); ^-inspired by python slicing. ...and .at(-2); will give you the next-to-last element – Nadu Dec 24 '22 at 09:58
106

Why so difficult?

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

Benjineer
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Quidn
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    Nicely done. Handles presence or absence of `#` and/or `?`. – Benjineer May 22 '16 at 08:54
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    This should be the top answer. – Henrik Petterson Mar 01 '17 at 13:22
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    @hayatbiralem gave a better answer on another comment : url.split('#').shift().split('?').shift().split('/').pop() You have to start by splitting on # and ? and afterwards split on / or you will catch the / parts of the anchor and querystring instead of the one from your path (which will typically lead to errors on amazon s3 protected urls ) – systho Jan 30 '18 at 09:22
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    What if URL doesn't actually has a filename in it? e.g: `http://www.example.com` this function will return "www.example.com" which is incorrect. It should be either empty string or null. – Slava Fomin II Jan 11 '19 at 12:10
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    Won't work for `"http://www.example.com/filenam.zip?passkey=1/2"` – tsh Nov 08 '19 at 07:33
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    @tsh (and Slava too) you are right, do note that other solutions works with those inputs as argument (like this one:https://stackoverflow.com/a/53560218/903998) but not this one. Although this was a good approach for reaching a solution. – Victor Jun 20 '20 at 12:32
  • Recently I realized that non-encoded slashes and question marks are explicitly allowed within the fragment due to the RFC 3986. So yes, this won't properly handle some legit URLs and I will rewrite the code later. Thanks for the inform. – Quidn Jun 22 '20 at 04:32
  • @tsh Use [the version by @hayatbiralem](https://stackoverflow.com/questions/511761/js-function-to-get-filename-from-url/48554885#comment61576914_17143667) – rinogo Apr 01 '21 at 15:59
  • Or, use the [super sexy version](https://stackoverflow.com/a/66908174/114558) I just posted! – rinogo Apr 01 '21 at 16:22
31

Use the match function.

function GetFilename(url)
{
   if (url)
   {
      var m = url.toString().match(/.*\/(.+?)\./);
      if (m && m.length > 1)
      {
         return m[1];
      }
   }
   return "";
}
David Morton
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16

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);
David Sherret
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brandonjp
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13

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}`);
sergdenisov
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    With new built-in Array method we can do `new URL(url).pathname.split('/').at(-1)`. `.pop()` still works tho. – Dale Ryan Mar 08 '23 at 04:52
12

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
}
Victor
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    Actually, this is the only solution working for all the edge cases I'm aware of. At least in the supported browsers (>89%). – Slava Fomin II Jan 11 '19 at 12:24
  • Not in my browser! **SyntaxError: Unexpected token ':'** –  Jun 10 '19 at 02:20
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    Any solution with URL doesn't work for relative URLs – Denis G. Labrecque Jul 18 '20 at 17:15
  • @DenisG.Labrecque you are right this solution doesn't works with "relative URL's" e.g.: `questions/511761/js-function-to-get-filename-from-url/53560218?noredirect=1#comment111357776_53560218` as using that string the URL object won't be constructed because of an invalid URL as argument. This solution will work with a fully qualified URL or "absolute", i mean, the url must begin with a protocol. In that case you may add an "http://" or ("whatever://") prefix to the relative URL (thus it becomes absolute) and this code will work as a charm. – Victor Aug 17 '20 at 13:15
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    Code golf fun: `f=u=>(p=>(i=>-1!==i?p.substring(i+1):p)(p.lastIndexOf('/')))(new URL(u).pathname);` – Mr. Polywhirl Sep 30 '20 at 12:40
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    You can even omit the `-1 !== index` check and just write `return pathname.substring(index + 1);` since -1 + 1 will return 0. – Robbendebiene Jul 20 '22 at 17:31
  • @Robbendebiene nice move!! updating snippet! – Victor Jul 21 '22 at 14:45
8

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.

Benjineer
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  • This is the best answer when using regex. – meYnot Apr 07 '16 at 10:44
  • What if URL doesn't actually has a filename in it? e.g: `http://www.example.com` this function will return "www.example.com" which is incorrect. It should be either empty string or null. – Slava Fomin II Jan 11 '19 at 12:12
6

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.

Denis G. Labrecque
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5

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)

Nadir
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  • Why do you expect to get "questions.html" when it is asked for get the name "questions" only? –  Jun 10 '19 at 01:57
  • Wrong. file extension is part of the name right? and it is asked filename right? – Nadir Jun 13 '19 at 11:36
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    See the second line in question: I need a javascript function to give me the 'th' value from that. –  Jun 14 '19 at 13:54
  • Ok, I missed it ;-) Then it is a onliner ie .split('.').pop() – Nadir Jun 15 '19 at 14:37
5

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')
tsh
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  • Why make it so complicated? It gets unsafe. –  Jun 16 '19 at 21:41
  • @PauliSudarshanTerho What do you mean by _unsafe_? Anything here is _unsafe_? – tsh Jun 17 '19 at 01:44
  • Simplicity = Safe. See my solution. Complexity = Unsafe. –  Jun 17 '19 at 02:27
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    @PauliSudarshanTerho But an answer should first be correct, then simple. And I can't see how simple = safe holds. – tsh Jun 17 '19 at 02:29
  • As you note some OS may have issues `URL()` and some browsers do not support it? Where is `filenameWithoutExtension()` used? –  Jun 17 '19 at 03:04
  • @PauliSudarshanTerho `URL` is browser specified, if you are using node.js, you may want some package which provide similar functions to it. Most recent browsers should support `URL`, but this may missing if you are working on old ones like IE. A filename contains special characters may not be valid on some OS, that's nothing to do with how to parse it (e.g. `http://www.example.com/a|b.txt` would be perfect valid URL. And `a|b.txt` would also be valid, but confusing, filename on Linux. But you cannot / should not name a file like this on Windows. – tsh Jun 17 '19 at 03:10
  • Sounds unsafe to me. How can I check `"https://www.example.com/"` or `"https://www.example.com"` give empty file name with your solution? –  Jun 17 '19 at 03:17
  • @PauliSudarshanTerho If so, you only need the first line, `decodeURIComponent(new URL('http://www.example.com').pathname.split('/').pop());` will give you empty string as you expected. – tsh Jun 17 '19 at 03:34
  • I remember your advise: *an answer should first be correct, then simple* - Very good! We are right on different things, but sure you are right.. often safety may require more code.. fore example check on null and fixes of all kinds to not break. Devil is in the details. –  Jun 17 '19 at 19:49
  • This should be marked as the correct answer. +1 for using URL and +99 for decodeURIComponent – Kunal Oct 26 '22 at 22:20
3

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 "";
}
James
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2

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.

rinogo
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    probably one of the best answers i've seen on SO... reason saying that is thorough and intuitive for testing – petrosmm Jun 10 '23 at 14:47
1

Using jQuery with the URL plugin:

var file = jQuery.url.attr("file");
var fileNoExt = file.replace(/\.(html|htm)$/, "");
// file == "th.html", fileNoExt = "th"
Chase Seibert
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1

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')
  })
})
cancerbero
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0

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"}

adam5990
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0

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);
}
HeavenCore
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  • True, but the importance of query solutions is questionable because it is not mentioned in the question. It is a choise between using '/' in query if any query at all or having shorter code. –  Jun 16 '19 at 21:30
0

Try this

url.substring(url.lastIndexOf('/')+1, url.length)
Abhishek Sharma
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0
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.

Karl.S
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    Doesn't split on '.' to remove the extension as asked for –  Jun 16 '19 at 21:32
  • @PauliSudarshanTerho you are right, it seems I did not read the question well, I corrected my answer. My code is less performant than the accepted answer, however it works with filenames like "news.inc.php" and it is easier to read. – Karl.S Jun 17 '19 at 18:22
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    The accepted answer is not correct at all: *Miss a lot of points.. does not check if url is undefined ... does not split on '.' ... does not remove eventual query string ...*. It is my comment there –  Jun 17 '19 at 19:37
0
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 :)

  • See my last point: *Not asked, but you can also have a query string without '/' and '.'* so that url has '/' in the query. –  Jun 17 '19 at 02:36
  • Between last '/' and last '.' is an empty name and I checked that gave it correctly ignoring the `.vimrc` extension. –  Jun 17 '19 at 02:44
  • It gives `%E6%96%87%E4%BB%B6%E5%90%8D`, shouldn't it? –  Jun 17 '19 at 03:37
  • I pasted the symbols into `http://example.com/文件名.html` and the `alert()` say `文件名` –  Jun 17 '19 at 03:50
0

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.

AmerllicA
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0

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);
MMMahdy-PAPION
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0

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.

carloswm85
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0
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

AlexWebLab
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-1

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

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