How to trim a file extension from a String in JavaScript?

Question

For example, assuming that x = filename.jpg, I want to get filename, where filename could be any file name (Let's assume the file name only contains [a-zA-Z0-9-_] to simplify.).

I saw x.substring(0, x.indexOf('.jpg')) on DZone Snippets, but wouldn't x.substring(0, x.length-4) perform better? Because, length is a property and doesn't do character checking whereas indexOf() is a function and does character checking.

Solution 1

If you know the length of the extension, you can use x.slice(0, -4) (where 4 is the three characters of the extension and the dot).

If you don't know the length @John Hartsock regex would be the right approach.

If you'd rather not use regular expressions, you can try this (less performant):

filename.split('.').slice(0, -1).join('.')

Note that it will fail on files without extension.

Solution 2

Not sure what would perform faster but this would be more reliable when it comes to extension like .jpeg or .html

x.replace(/\.[^/.]+$/, "")

Solution 3

In node.js, the name of the file without the extension can be obtained as follows.

const path = require('path');
const filename = 'hello.html';
    
path.parse(filename).name;     //=> "hello"
path.parse(filename).ext;      //=> ".html"
path.parse(filename).base; //=> "hello.html"

Further explanation at Node.js documentation page.

Solution 4

x.length-4 only accounts for extensions of 3 characters. What if you have filename.jpegor filename.pl?

EDIT:

To answer... sure, if you always have an extension of .jpg, x.length-4 would work just fine.

However, if you don't know the length of your extension, any of a number of solutions are better/more robust.

x = x.replace(/\..+$/, '');

OR

x = x.substring(0, x.lastIndexOf('.'));

OR

x = x.replace(/(.*)\.(.*?)$/, "$1");

OR (with the assumption filename only has one dot)

parts = x.match(/[^\.]+/);
x = parts[0];

OR (also with only one dot)

parts = x.split(".");
x = parts[0];

Solution 5

You can perhaps use the assumption that the last dot will be the extension delimiter.

var x = 'filename.jpg';
var f = x.substr(0, x.lastIndexOf('.'));

If file has no extension, it will return empty string. To fix that use this function

function removeExtension(filename){
    var lastDotPosition = filename.lastIndexOf(".");
    if (lastDotPosition === -1) return filename;
    else return filename.substr(0, lastDotPosition);
}

View more solutions

Comments

[gsnedders]

You probably want to also disallow / as a path separator, so the regexp is /\.[^/.]+$/

[ma11hew28]

I like this solution the best. It's clean, and I can use it cause I know the file extension is always .jpg. I was looking for something like Ruby's x[0..-5], and x.slice(0, -4) looks great! Thanks! And thank you to everyone else for all the other robust alternatives provided!

[Roko C. Buljan]

?? You can have a filename ex: "summer.family.jpg" in that case split('.')[0] will return only a partial file name. I would remove that one from the answer, or clearly state underneath the issue for that example. @basarat ...

[radicand]

Something I do frequently regarding part splits: var parts = full_file.split("."); var ext = parts[parts.length-1]; var file = parts.splice(0,parts.length-1).join(".");

[Brad]

Warning, this fails if there happens to be no filename extension. You're left with an empty string.

[bunjeeb]

this is not the optimal solution, please check other solutions below.

[basic6]

And if you're not 100% sure about the length of the extension, then don't this: "picture.jpeg".slice(0, -4) -> "picture."

[Jonathan Rowny]

Shorter version that accounts for no dots. var f = x.substr(0, x.lastIndexOf('.')) || x; This works because an empty string is falsy, therefore it returns x.

[Steve Seeger]

This works for any length of file extension (.txt or .html or .htaccess) and also allows for the file name to contain additional period (.) characters. It wouldn't handle eg .tar.gz due to the extension itself containing a period. It's more common for a file name to contain additional periods than a file extension. Thanks!

[T J]

"If you're going for performance, don't go for javascript at all" - What else are you suggesting to use in web applications..?

[Lee Brindley]

x.split(".") should not even be considered an answer. I know I use a '.' in almost all of my file naming conventions, i.e. 'survey.controller.js', or 'my.family.jpg'.

[Jeff B]

@Lee2808: Hence the warning of only one dot. This is simply meant to show that there are a number of approaches, depending on the application. I would certainly use one of the other methods in almost all cases.

[rossanmol]

This is dangerous solution, cause you don't really know the length of the format.

[Lucas Moeskops]

He doesn't mention web applications.

[T J]

This question was asked and answer was posted in 2010, 7 years ago, and JavaScript was pretty much used only in web applications. (Node was just born, it didn't even had a guide or NPM at that time)

[Lucas Moeskops]

;-) Still, if performance matters on tasks like this, you might consider doing this on the backend and process the results on the frontend.

[Dziad Borowy]

x = x.substr(0, x.lastIndexOf('.')); - you probably meant x = x.substring(0, x.lastIndexOf('.'));?

[Jeff B]

@tborychowski When the first parameter is 0, substr and substring do the same thing. See: stackoverflow.com/questions/3745515/…

[Dziad Borowy]

"it works for this case" isn't always the same as "it's correct implementation" :-) substring's second parameter is indexEnd, substr takes length, so IMO substring fits better here.

[Jeff B]

Ah, I was actually about to argue the same point until I realized that, yes, I have them backwards. So, yes, fair point. Why javascript uses 2 functions that are easily confused is beyond me.

[Evil]

It should not be name.join('') but name.join('.'). You split by dot but join by comma, so hello.name.txt returns hello, name

[Steven]

@Vik There's a difference between the 'right answer' and the accepted answer. An accepted answer is just the answer that was helpful for the one who asked the question.

[Asif Ali]

fails with hello.tar.gz, output is hellotar.

[Little Brain]

#AsifAli thanks you are right, I forgot to readd the file extension. I've updated the answer, I hope it works now.

[Alex Chuev]

I suppose that there may be issues with the Windows platform because there can be back slashes. So the regexp should be /\.[^/\\.]+$/.

[Leonid Zadorozhnykh]

x = x.substring(0, x.lastIndexOf('.')); is my choice. Thank you!

[nollaf126]

@John Hartsock, I'm trying to use this with Adobe Extendscript and it's giving me a syntax error message and highlighting the ]+$ part. I think the ] is causing the problem. Does anyone know a workaround for this? I'm using it to append a string to the end of a file name before its extension so I can export graphics from an InDesign document.

[John Hartsock]

@nollaf126 sounds like you need to escape the ]

[Qian Chen]

What does [^/.] mean?

[John Hartsock]

@ElgsQianChen here is a great tool for you to help answer your question regexr.com

[nfdavenport]

@ElgsQianChen the stackoverflow answer you found to your problem uses regex, now you have two problems. regex.info/blog/2006-09-15/247

[Charlie]

This answer is pretty restricted to server-side node. If you try to use this in react code, it doesn't seem to import.

[sboss]

Does not handle cases where the string does not contain an extension.

[mixdev]

or path.split('.').pop() for one part file extensions

[Munim]

He was actually trying to get the file name, not the extension!

[Jespertheend]

if you want to remove an extension from a path including the directories, you can do var parsed = path.parse(filename) followed by path.join(parsed.dir, parsed.name).

[Ciro Santilli OurBigBook.com]

This is the only method so far that works for full paths: path/name.ext -> paht/name instead of just returning name, but I would rather do with with fs.parse although it is a bit more verbose: stackoverflow.com/a/59576950/895245

[bicarlsen]

Another possibility is let base = path.basename( file_path, path.extname( file_path ) ).

[Dan Dascalescu]

The path.parse answer is simpler.

[Alexander]

"If you know the length of the extension" It's been decades since this was an acceptable assumption to make. Don't use this anymore.

[gsaandy]

filename.split(".").shift();

[Donald Duck]

While this code may answer the question, providing additional context regarding why and/or how this code answers the question improves its long-term value.

[Nick Grealy]

I like this answer... to add to it: If you know the extension beforehand (or if the extension is a variable, then I find it more readable to say: filename.slice(0, -'.zip'.length) or filename.slice(0, -extension.length).

[OOPS Studio]

@Alexander Sorry to reply so late, but this is a perfectly acceptable solution in situations where you know the length of the extension, and those situations DO exist despite what you may think. In my case I'm filtering out all files that don't end with ".js" so it's not like I'm going to somehow get a different extension length anyway.

[Ratul Hasan]

Not working on files without extension. gave a hard time detecting this bug at work.

[Felix Liu]

filename.split('.').slice(0, -1).join('.') || filename works even for files without extension.

[Angel]

I think this one is the best because it's really easy to understand