Extract File Extension from File Path

Extract file extension from file path

This is the sort of thing that easily found with R basic tools. E.g.: ??path.

Anyway, load the tools package and read ?file_ext .

Extracting extension from filename in Python

Use os.path.splitext:

>>> import os
>>> filename, file_extension = os.path.splitext('/path/to/somefile.ext')
>>> filename
'/path/to/somefile'
>>> file_extension
'.ext'

Unlike most manual string-splitting attempts, os.path.splitext will correctly treat /a/b.c/d as having no extension instead of having extension .c/d, and it will treat .bashrc as having no extension instead of having extension .bashrc:

>>> os.path.splitext('/a/b.c/d')
('/a/b.c/d', '')
>>> os.path.splitext('.bashrc')
('.bashrc', '')

How to get the File Extension from a string Path

You can use the extension function in the path package to get the extension from a file path:

import 'package:path/path.dart' as p;

final path = '/some/path/to/file/file.dart';

final extension = p.extension(path); // '.dart'

If your file has multiple extensions, like file.dart.js, you can specify the optional level parameter:

final extension = p.extension('file.dart.js', 2); // '.dart.js'

How can I get file extensions with JavaScript?

Newer Edit: Lots of things have changed since this question was initially posted - there's a lot of really good information in wallacer's revised answer as well as VisioN's excellent breakdown

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Edit: Just because this is the accepted answer; wallacer's answer is indeed much better:

return filename.split('.').pop();

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My old answer:

return /[^.]+$/.exec(filename);

Should do it.

Edit: In response to PhiLho's comment, use something like:

return (/[.]/.exec(filename)) ? /[^.]+$/.exec(filename) : undefined;

Extract filename and extension in Bash

First, get file name without the path:

filename=$(basename -- "$fullfile")
extension="${filename##*.}"
filename="${filename%.*}"

Alternatively, you can focus on the last '/' of the path instead of the '.' which should work even if you have unpredictable file extensions:

filename="${fullfile##*/}"

You may want to check the documentation :

• On the web at section "3.5.3 Shell Parameter Expansion"
• In the bash manpage at section called "Parameter Expansion"

How to find the extension of a file in C#?

Path.GetExtension

string myFilePath = @"C:\MyFile.txt";
string ext = Path.GetExtension(myFilePath);
// ext would be ".txt"

How do I get the file extension of a file in Java?

In this case, use FilenameUtils.getExtension from Apache Commons IO

Here is an example of how to use it (you may specify either full path or just file name):

import org.apache.commons.io.FilenameUtils;

// ...

String ext1 = FilenameUtils.getExtension("/path/to/file/foo.txt"); // returns "txt"
String ext2 = FilenameUtils.getExtension("bar.exe"); // returns "exe"

Maven dependency:

<dependency>
  <groupId>commons-io</groupId>
  <artifactId>commons-io</artifactId>
  <version>2.6</version>
</dependency>

Gradle Groovy DSL

implementation 'commons-io:commons-io:2.6'

Gradle Kotlin DSL

implementation("commons-io:commons-io:2.6")

Others https://search.maven.org/artifact/commons-io/commons-io/2.6/jar

Remove file extension from a file name string

I used the below, less code

string fileName = "C:\file.docx";
MessageBox.Show(Path.Combine(Path.GetDirectoryName(fileName),Path.GetFileNameWithoutExtension(fileName)));

Output will be

C:\file

Extracting a file extension from a given path in Rust idiomatically

In idiomatic Rust the return type of a function that can fail should be an Option or a Result. In general, functions should also accept slices instead of Strings and only create a new String where necessary. This reduces excessive copying and heap allocations.

You can use the provided extension() method and then convert the resulting OsStr to a &str:

use std::path::Path;
use std::ffi::OsStr;

fn get_extension_from_filename(filename: &str) -> Option<&str> {
    Path::new(filename)
        .extension()
        .and_then(OsStr::to_str)
}

assert_eq!(get_extension_from_filename("abc.gz"), Some("gz"));

Using and_then is convenient here because it means you don't have to unwrap the Option<&OsStr> returned by extension() and deal with the possibility of it being None before calling to_str. I also could have used a lambda |s| s.to_str() instead of OsStr::to_str - it might be a matter of preference or opinion as to which is more idiomatic.

Notice that both the argument &str and the return value are references to the original string slice created for the assertion. The returned slice cannot outlive the original slice that it is referencing, so you may need to create an owned String from this result if you need it to last longer.

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