How can I extract the filename without extention from the following file path:
D:\Projects\Extract\downtown - second.pdf
The following regular expression gives me the filename with extention: [^\\]*$
e.g. downtown - second.pdf
The following regular expression gives me the filename without extention: (.+)(?=(\.))
e.g. D:\Projects\Extract\downtown - second
I'm struggling to combine the two into one regular expression to give me the results I want: downtown - second
I suspect that your 2nd regex would not give you the output you have shown. It will give you the complete string till the first period (.).
To get just the file name without extension, you can use this regex: -
[^\\]*(?=[.][a-zA-Z]+$)
I have just replaced (.+) in your 2nd regex with the [^\\]* from your first regex, and added pattern to match pdf till the end.
Now this pattern will match 0 or more repetition of any character but backslash(\), followed by a . and then 1 or more repetition of alphabets making up extension.
I made up this one, which allows to capture most of the possibilities:
/[^\\\/]+(?=\.[\w]+$)|[^\\\/]+$/
/path/to/file
/path/to/file.txt
/path.with/dots.to/file.txt
/path/to/file.with.dots.txt
file.txt
C:\path\to\file.txt
and so on...
I captured file from /path/to/file.pdf by using following regex:
[^/]*(?=\.[^.]+($|\?))
Hope this helps you
I had to use an extra backslash before the first ']' to make this work
[^\\\]*(?=[.][a-zA-Z]+$)
I use this pattern
[^\/]+[.+\.].*$ for / path separator
[^\\]+[.+\.].*$ for \ path separator
hich matches the filename at the end of the string without worrying about characters. There is one exception that if the path for some reason has a folder with a period in it this will get upset. Linux hidden directories that are preceded with a . like .rvm are unaffected.
Hope this helps.
http://rubular.com/r/LNrI4inMU1
Related
I don't know anything about Notepad++ Regex.
This is the data I have in my CSV:
6454345|User1-2ds3|62562012032|324|148|9c1fe63ccd3ab234892beaf71f022be2e06b6cd1
3305611|User2-42g563dgsdbf|22023001345|0|0|c36dedfa12634e33ca8bc0ef4703c92b73d9c433
8749412|User3-9|xgs|f|98906504456|1534|51564|411b0fdf54fe29745897288c6ad699f7be30f389
How can I use a Regex to remove the 5th and 6th column? The numbers in the 5th and 6th column are variable in length.
Another problem is the User row can also contain a |, to make it even worse.
I can use a macro to fix this, but the file is a few millions lines long.
This is the final result I want to achieve:
6454345|User1-2ds3|62562012032|9c1fe63ccd3ab234892beaf71f022be2e06b6cd1
3305611|User2-42g563dgsdbf|22023001345|c36dedfa12634e33ca8bc0ef4703c92b73d9c433
8749412|User3-9|xgs|f|98906504456|411b0fdf54fe29745897288c6ad699f7be30f389
I am open for suggestions on how to do this with another program, command line utility, either Linux or Windows.
Match \|[^|]+\|[^|]+(\|[^|]+$)
Repalce $1
Basically, Anchor to the end of the line, and remove columns [-1] and [-2] (I assume columns can't be empty. Replace + with * if they can)
If you need finer detail then that, I'd recommend writing a Java or Python script to manual parse and rewrite the file for you.
I've captured three groups and given them names. If you use a replace utility like sed or vimregex, you can replace remove with nothing. Or you can use a programming language to concatenate keep_before and keep_after for the desired result.
^(?<keep_before>(?:[^|]+\|){3})(?<remove>(?:[^|]+\|){2})(?<keep_after>.*)$
You may have to remove the group namings and use \1 etc. instead, depending on what environment you use.
Demo
From Notepad++ hit ctrl + h then enter the following in the dialog:
Find what: \|\d+\|\d+(\|[0-9a-z]+)$
Replace with: $1
Search mode: Regular Expression
Click replace and done.
Regex Explain:
\|\d+ : match 1st string that starts with | followed by number
\|\d+ : match 2nd string that starts with | followed by number
(\|[0-9a-z]+): match and capture the string after the 2nd number.
$ : This is will force regex search to match the end of the string.
Replacement:
$1 : replace the found string with whatever we have between the captured group which is whatever we have between the parentheses (\|[0-9a-z]+)
I've got this file here:
\\prdflsrvcl2.unicreprd.local\Integracao-PRD\GestaoTangiveis\APD\FW_A_enviar_correio_electronico_Imagem_(384).jpg-1-52FFN8.msg
With the regular expression, I want this part: FW_A_enviar_correio_electronico_Imagem_(384).jpg-1-52FFN8.msg
I'm using the following regex:
[[:alnum:]-_]+\.[a-zA-Z]*$
However, I get .jpg-1-52FFN8.msg instead of what I want.
However if the file name was without the dot before the jpg, I would get FW_A_enviar_correio_electronico_Imagem_(384)jpg-1-52FFN8.msg instead.
Basically, I want the filename with the extension.
Thanks.
Just try with following regex:
[^\\]+$
It will match all characters from the end to the first occurence of \.
Example: https://regex101.com/r/eJ8zG2/1
You can use the following regex and refer the first capturing group which is (.*):
/(?!.*\\)(.*)/g
Match all things which is not followed by \
Example
I have some lines of code I am trying to remove some leading text from which appears like so:
Line 1: myApp.name;
Line 2: myApp.version
Line 3: myApp.defaults, myApp.numbers;
I am trying and trying to find a regex that will remove anything up to (but excluding) myApp.
I have tried various regular expressions, but they all seem to fail when it comes to line 3 (because myApp appears twice).
The closest I have come so far is:
.*?myApp
Pretty simple - but that matches both instances of myApp occurrences in Line 3 - whereas I'd like it to match only the first.
There's a few hundred lines - otherwise I'd have deleted them all manually by now.
Can somebody help me? Thanks.
You need to add an anchor ^ which matches the starting point of a line ,
^.*?(myApp)
DEMO
Use the above regex and replace the matched characters with $1 or \1. So that you could get the string myApp in the final result after replacement.
Pattern explanation:
^ Start of a line.
.*?(myApp) Shortest possible match upto the first myApp. The string myApp was captured and stored into a group.(group 1)
All matched characters are replaced with the chars present inside the group 1.
Your regular expression works in Perl if you add the ^ to ensure that you only match the beginnings of lines:
cat /tmp/test.txt | perl -pe 's/^.*?myApp/myApp/g'
myApp.name;
myApp.version
myApp.defaults, myApp.numbers;
If you wanted to get fancy, you could put the "myApp" into a group that doesn't get captured as part of the expression using (?=) syntax. That way it doesn't have to be replaced back in.
cat /tmp/test.txt | perl -pe 's/^.*?(?=myApp)//g'
myApp.name;
myApp.version
myApp.defaults, myApp.numbers;
I need to extract just the filename (no file extension) from the following path....
\\my-local-server\path\to\this_file may_contain-any&character.pdf
I've tried several things, most based off of something like http://regexr.com?302m5 but can't quite get there
^\\(.+\\)*(.+)\.(.+)$
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 "(.+)$" matches extension.
This will get the filename but will also get the dot. You might want to truncate the last digit from it in your code.
[\w-]+\.
Update
#Geoman if you have spaces in file name then use the modified pattern below
[ \w-]+\. (space added in brackets)
Demo
This is just a slight variation on #hmd's so you don't have to truncate the .
[ \w-]+?(?=\.)
Demo
Really, thanks goes to #hmd. I've only slightly improved on it.
Try this:
[^\\]+(?=\.pdf$)
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 $1, or \1, or the library will provide some method for getting capturing group by its number after regexp match.
I use #"[^\\]+$"
That gives the filename including the extension.
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+$)/
console.log('/path/to/file.png'.match(regex))
console.log('/path/to/video.webm'.match(regex))
console.log('/path/to/weird.file.gif'.match(regex))
console.log('/path with/spaces/and file.with.spaces'.match(regex))
If anyone is looking for a windows absolute path (and relative path) javascript regular expression in javascript for files:
var path = "c:\\my-long\\path_directory\\file.html";
((/(\w?\:?\\?[\w\-_\\]*\\+)([\w-_]+)(\.[\w-_]+)/gi).exec(path);
Output is:
[
"c:\my-long\path_directory\file.html",
"c:\my-long\path_directory\",
"file",
".html"
]
Here's a slight modification to Angelo's excellent answer that allows for spaces in the path, filename and extension as well as missing parts:
function parsePath (path) {
var parts = (/(\w?\:?\\?[\w\-_ \\]*\\+)?([\w-_ ]+)?(\.[\w-_ ]+)?/gi).exec(path);
return {
path: parts[0] || "",
folder: parts[1] || "",
name: parts[2] || "",
extension: parts[3] || "",
};
}
If you want to return the file name with its extension, Regex should be as below:
[A-Za-z0-9_\-\.]+\.[A-Za-z0-9]+$
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 to return file name without the last extension Regex should be as below:
[A-Za-z0-9_\-\.]+(?=\.[A-Za-z0-9]+$)
Which returns full file name without last extension(eg: "filename" for "filename.some" and "filename.some" for "filename.some.other")
Click the Explain button on these links shown TEST to see how they work.
This is specific to the pdf extension.
TEST ^.+\\([^.]+)\.pdf$
This is specific to any extension, not just pdf.
TEST ^.+\\([^.]+)\.[^\.]+$
([^.]+)
This is the $1 capture group to extract the filename without the extension.
\\my-local-server\path\to\this_file may_contain-any&character.pdf
will return
this_file may_contain-any&character
TEST ^(.*[\\\/])?(.*?)(\.[^.]*?|)$
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 $3 capture group is the extension (only the last)
works for:
C:\folder1\folder2\foo.ext
C:\folder1\folder2\foo.ext1.ext
C:\folder1\folder2\name-without extension
only name
name.ext
C:\folder1\folder2\foo.ext
/folder1/folder2/foo.ext
C:\folder1\folder2\foo
C:\folder1\folder2\
C:\special&chars\folder2\f [oo].ext1.e-x-t
Answer with:
File name and directory space support
Named capture group
Gets unlimited file extensions (captures file.tar.gz, not just file.tar)
*NIX and Win support
^.+(\\|\/)(?<file_name>([^\\\/\n]+)(\.)?[^\n\.]+)$
Explanation:
^.+(\\|\/) Gets anything up to the final / or \ in a file path
(?<file_name> Begin named capture group
([^\\\/\n]+) get anything except for a newline or new file
(\.)?[^\n\.]+ Not really needed but it works well for issues with odd characters in file names
)$ 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\.]+)$"
Here is an alternative that works on windows/unix:
"^(([A-Z]:)?[\.]?[\\{1,2}/]?.*[\\{1,2}/])*(.+)\.(.+)"
First block: path
Second block: dummy
Third block: file name
Fourth block: extension
Tested on:
".\var\www\www.example.com\index.php"
"\var\www\www.example.com\index.php"
"/var/www/www.example.com/index.php"
"./var/www/www.example.com/index.php"
"C:/var/www/www.example.com/index.php"
"D:/var/www/www.example.com/index.php"
"D:\\var\\www\\www.example.com\\index.php"
"\index.php"
"./index.php"
This regular expression extract the file extension, if group 3 isn't null it's the extension.
.*\\(.*\.(.+)|.*$)
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`
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}.*)""","$3")
Direct approach:
To answer your question as it's written, this will provide the most exact match:
^\\\\my-local-server\\path\\to\\(.+)\.pdf$
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.
Does this work...
.*\/(.+)$
Posting here so I can get feedback
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-. ]+(?=[.])
try this
[^\\]+$
you can also add extension for specificity
[^\\]+pdf$
If I want to match only fileName, i.e,
in C://Directory/FileName.cs, somehow ignore everything before FileName.cs using Regex.
How can I do it?
I need this for a Compiled UI I am working on ... can't use programming language as it only accepts Regex.
Any ideas?
Something like this might work:
[^/]*$
It matches all characters to the end of the line that are not "/"..
If you want to match paths that use the "\" path separator you would change the regex to:
[^\]*$
But do make sure to escape the "\" character if your programming language or environment requires it. For instance you might have to write something like this:
[^\\]*$
EDIT
I removed the leading "/" and trailing "/" as they may be confusing since they are not really part of the regEx but they are very common of representing a regular expression.
And of course, depending on the features that the regEx engine supports you may be able to use look-ahead/look-behind and capturing to craft a better regEx.
What language are you using? Why are you not using the standard path mechanisms of that language?
How about http://msdn.microsoft.com/en-us/library/system.io.path.aspx ?
Based on your comment of needing to exclude paths that do not match 'abc', try this:
^.+/(?:(?!abc)[^/])+$
Completely split out in regex comment mode, that is:
(?x) # flag to enable comments
^ # start of line
.+ # match any character (except newline)
# greedily one or more times
/ # a literal slash character
(?: # begin non-capturing group
(?! # begin negative lookahead
# (contents must not appear after the current position)
abc # literal text abc
) # end negative lookahead
[^/] # any character that is not a slash
) # end non-capturing group
+ # repeat the above nc group one or more times
# (essentially, we keep looking for non-backspaces that are not 'abc')
$ # end of line
The regex expression that did it for me was
[^\/]*$
I'm way late to the party and I'm also ignoring the requirement of regex because, as J-16 SDiZ pointed out, sometimes there is a better solution. Even though the question is 4 years old, people looking for a simple solution deserve choices.
Try using the following:
public string ConvertFileName(string filename)
{
string[] temparray = filename.Split('\\');
filename = temparray[temparray.Length - 1];
return filename;
}
This method splits the string on the "\" character, stores the resulting strings in an array and returns the last element of the array (the filename).
Though the OP seems to be writing for UNIX, it doesn't take much to figure out how to tailor it to your particular need.
Seeing as filename can be interpreted as the basename by some. Then, this example can extract the filename/basename for any files that may not have an extension for some reason. It can also get the last directory in the same fashion.
You can see how it works and test it here.
https://regexr.com/4ht5v
The regexp is:
.+?\\(?=\w+)|\.\w+$|\\$
Before:
C:\Directory\BaseFileName.ext
C:\Directory\BaseFileName
C:\This is a Directory\Last Directory With trailing backslash\
C:\This is a Directory\Last Directory Without trailing backslash
After:
BaseFileName
BaseFileName
Last Directory With trailing backslash
Last Directory Without trailing backslash
For the sake of completion, this is how it would work with JavaScript should anyone require it.
// Example of getting a BaseFileName from a path
var path = "C:\\Directory\\FileName.cs";
var result = path.replace(/.+?\\(?=\w+)|\.\w+$|\\$/gm,"");
console.log(result);
Try this (working with / and \):
[^\/|\\]*$
I would use: ./(.$)
The parenthesis mark a group wich is the file name.
The regular expression you use may vary dependig on the regex syntax(PCRE, POSIX)
I sugest you use a regex tool, there are several for windows and linux:
Windows - http://sourceforge.net/projects/regexcreator/
Windows - http://weitz.de/regex-coach/
Linux - kodos
Hope it helps
just a variation on miky's that works for both filesystem path characters:
[^\\/]*\s
Suppose the file name has special characters, specially when supporting MAC where special characters are allowing in filenames, server side Path.GetFileName(fileName) fails and throws error because of illegal characters in path. The following code using regex come for the rescue.
The following regex take care of 2 things
In IE, when file is uploaded, the file path contains folders aswell (i.e. c:\samplefolder\subfolder\sample.xls). Expression below will replace all folders with empty string and retain the file name
When used in Mac, filename is the only thing supplied as its safari browser and allows special chars in file name.
var regExpDir = #"(^[\w]:\\)([\w].+\w\\)";
var fileName = Regex.Replace(fileName, regExpDir, string.Empty);
I did this without RegEx in Powershell:
Put the link in a variable
$Link = "http://some.url/some/path/file.name"
Split the link on the "/" character
$split = $Link.Split("/")
Count the splits
$SplitCount = $Split.Count
Target the filename
$Split[$SplitCount -1]
Full code :
$Link = "http://some.url/some/path/file.name"
$Split = $Link.Split("/")
$SplitCount = $Split.Count
$Split[$SplitCount -1]
A rather elegant solution with lookahead and lookbehind wasn't mentioned:
(?<=.+)(?=.cs)