Regex for extracting filename from path

Question

I need to extract just the filename (no file extension) from the following path....

\\\\my-local-server\\path\\to\\this_file may_contain-any&character.pdf

I've tried several things, most based off of something like http://regexr.com?302m5 but can't quite get there

Answer 1

^\\(.+\\)*(.+)\.(.+)$

This regex has been tested on these two examples:

\\var\\www\\www.example.com\\index.php
\\index.php

First block "(.+\\)*" matches directory path.
Second block "(.+)" matches file name without extension.
Third block "(.+)$" matches extension.

Answer 2

This will get the filename but will also get the dot. You might want to truncate the last digit from it in your code.

[\w-]+\.

Update

@Geoman if you have spaces in file name then use the modified pattern below

[ \w-]+\.      (space added in brackets)

Demo

Answer 3

This is just a slight variation on @hmd's so you don't have to truncate the .

[ \w-]+?(?=\.)

Demo

Really, thanks goes to @hmd. I've only slightly improved on it.

Answer 4

Try this :

[^\\]+(?=\.pdf$)

It matches everything except back-slash followed by .pdf at the end of the string.

You can also (and maybe it's even better) take the part you want into the capturing group like that:

([^\\]+)\.pdf$

But how you refer to this group (the part in parenthesis) depends on the language or regexp flavor you're using. In most cases it'll be smth like $1 , or \\1 , or the library will provide some method for getting capturing group by its number after regexp match.

Answer 5

我使用@"[^\\\\]+$"这给出了包括扩展名的文件名。

Answer 6

If anyone is looking for a windows absolute path (and relative path) javascript regular expression in javascript for files:

var path = "c:\\my-long\\path_directory\\file.html";


((/(\w?\:?\\?[\w\-_\\]*\\+)([\w-_]+)(\.[\w-_]+)/gi).exec(path);

Output is:

[
"c:\my-long\path_directory\file.html", 
"c:\my-long\path_directory\", 
"file", 
".html"
]

Answer 7

Here's a slight modification to Angelo's excellent answer that allows for spaces in the path, filename and extension as well as missing parts:

function parsePath (path) {
    var parts = (/(\w?\:?\\?[\w\-_ \\]*\\+)?([\w-_ ]+)?(\.[\w-_ ]+)?/gi).exec(path);
    return {
        path: parts[0] || "",
        folder: parts[1] || "",
        name: parts[2] || "",
        extension: parts[3] || "",
    };
}

Answer 8

Here is an alternative that works on windows/unix:

"^(([AZ]:)?[\\.]?[\\\\{1,2}/]?.*[\\\\{1,2}/])*(.+)\\.(.+)"

First block: path
Second block: dummy
Third block: file name
Fourth block: extension

Tested on:

".\var\www\www.example.com\index.php"
"\var\www\www.example.com\index.php"
"/var/www/www.example.com/index.php"
"./var/www/www.example.com/index.php"
"C:/var/www/www.example.com/index.php"
"D:/var/www/www.example.com/index.php"
"D:\\var\\www\\www.example.com\\index.php"
"\index.php"
"./index.php"

Answer 9

Click the Explain button on these links shown TEST to see how they work.

---

This is specific to the pdf extension.

TEST ^.+\\\\([^.]+)\\.pdf$

---

This is specific to any extension, not just pdf .

TEST ^.+\\\\([^.]+)\\.[^\\.]+$

---

([^.]+) This is the $1 capture group to extract the filename without the extension .

---

\\\\my-local-server\\path\\to\\this_file may_contain-any&character.pdf

will return

this_file may_contain-any&character

Answer 10

I'm using this regex to replace the filename of the file with index . It matches a contiguous string of characters that doesn't contain a slash and is followed by a . and a string of word characters at the end of the string. It will retrieve the filename including spaces and dots but will ignore the full file extension.

 const regex = /[^\\\\/]+?(?=\\.\\w+$)/ console.log('/path/to/file.png'.match(regex)) console.log('/path/to/video.webm'.match(regex)) console.log('/path/to/weird.file.gif'.match(regex)) console.log('/path with/spaces/and file.with.spaces'.match(regex))

Answer 11

Answer with:

• File name and directory space support
• Named capture group
• Gets unlimited file extensions (captures file.tar.gz , not just file.tar )
• *NIX and Win support

^.+(\\\\|\\/)(?<file_name>([^\\\\\\/\\n]+)(\\.)?[^\\n\\.]+)$

Explanation:

1. ^.+(\\\\|\\/) Gets anything up to the final / or \\ in a file path
2. (?<file_name> Begin named capture group
3. ([^\\\\\\/\\n]+) get anything except for a newline or new file
4. (\\.)?[^\\n\\.]+ Not really needed but it works well for issues with odd characters in file names
5. )$ End named capture group and end line

Note that if you're putting this in a string and you need to escape backslashes (such as with C) you'll be using this string:

"^.+(\\\\\\\\|\\/)(?<file_name>([^\\\\\\/\\n]+)(\\.)?[^\\n\\.]+)$"

Answer 12

If you want return file name with it's extension Regex should be as bellow:

[A-Za-z0-9_\\-\\.]+\\.[A-Za-z0-9]+$

works for

path/to/your/filename.some
path/to/your/filename.some.other
path\to\your\filename.some
path\to\your\filename.some.other
http://path/to/your/filename.some
http://path/to/your/filename.some.other
And so on

Which returns full file name with extension(eg: filename.some or filename.some.other)

---

If you want return file name without last extension Regex should be as bellow:

[A-Za-z0-9_\-\.]+(?=\.[A-Za-z0-9]+$)

Which returns full file name without last extension(eg: "filename" for "filename.some" and "filename.some" for "filename.some.other")

Answer 13

此正则表达式提取文件扩展名，如果第 3 组不为空，则为扩展名。

.*\\(.*\.(.+)|.*$)

Answer 14

also one more for file in dir and root

   ^(.*\\)?(.*)(\..*)$

for file in dir

Full match  0-17    `\path\to\file.ext`
Group 1.    0-9 `\path\to\`
Group 2.    9-13    `file`
Group 3.    13-17   `.ext`

for file in root

Full match  0-8 `file.ext`
Group 2.    0-4 `file`
Group 3.    4-8 `.ext`

Answer 15

For most of the cases ( that is some win , unx path , separator , bare file name , dot , file extension ) the following one is enough:

 // grap the dir part (1), the dir sep(2) , the bare file name (3) 
 path.replaceAll("""^(.*)[\\|\/](.*)([.]{1}.*)""","$3")

Answer 16

Direct approach:

To answer your question as it's written, this will provide the most exact match:

^\\\\my-local-server\\path\\to\\(.+)\.pdf$

General approach:

This regex is short and simple, matches any filename in any folder (with or without extension) on both windows and *NIX:

.*[\\/]([^.]+)

If a file has multiple dots in its name, the above regex will capture the filename up to the first dot. This can easily be modified to match until the last dot if you know that you will not have files without extensions or that you will not have a path with dots in it.

If you know that the folder will only contain .pdf files or you are only interested in .pdf files and also know that the extension will never be misspelled, I would use this regex:

.*[\\/](.+)\.pdf$

Explanation:

• . matches anything except line terminators.
• * repeats the previous match from zero to as many times as possible.
• [\\\\/] matches a the last backslash or forward slash (previous ones are consumed by .* ). It is possible to omit either the backslash or the forward slash if you know that only one type of environment will be used. If you want to capture the path, surround .* or .*[\\\\/] in parenthesis.
• Parenthesis will capture what is matched inside them.
• [^.] matches anything that is not a literal dot.
• + repeats the previous match one or more times, as many as possible.
• \\. matches a literal dot.
• pdf matches the string pdf.
• $ asserts the end of the string.

If you want to match files with zero, one or multiple dots in their names placed in a variable path which also may contain dots, it will start to get ugly. I have not provided an answer for this scenario as I think it is unlikely.

Edit: To also capture filenames without a path, replace the first part with (?:.*[\\\\/])? , which is an optional non-capturing group.

Answer 17

Does this work...

.*\/(.+)$

Posting here so I can get feedback

Answer 18

Here a solution to extract the file name without the dot of the extension. I begin with the answer from @Hammad Khan and add the dot in the search character. So, dots can be part of the file name:

[ \w-.]+\.

Then use the regex look ahead (?= ) for a dot, so it will stop the search at the last dot (the dot before the extension), and the dot will not appears in the result:

[ \w-.]+(?=[.])

reorder, it's not necessary but look better:

[\w-. ]+(?=[.])

Answer 19

TEST ^(.*[\\\\\\/])?(.*?)(\\.[^.]*?|)$

example:

/^(.*[\\\/])?(.*?)(\.[^.]*?|)$/.exec("C:\\folder1\\folder2\\foo.ext1.ext")

result:

0: "C:\folder1\folder2\foo.ext1.ext"
1: "C:\folder1\folder2\"
2: "foo.ext1"
3: ".ext"

the $1 capture group is the folder
the $2 capture group is the name without extension
the $3 capture group is the extension (only the last)

works for:

• C:\\folder1\\folder2\\foo.ext
• C:\\folder1\\folder2\\foo.ext1.ext
• C:\\folder1\\folder2\\name-without extension
• only name
• name.ext
• C:\\folder1\\folder2\\foo.ext
• /folder1/folder2/foo.ext
• C:\\folder1\\folder2\\foo
• C:\\folder1\\folder2\\
• C:\\special&chars\\folder2\\f [oo].ext1.ext