I tried with [^0-9].* and [^\d].*
But none of them is working:
I only want to get filename which doesn't contain any numbers, in above case,
I need to get BUILDING.txt.
I also tried with
But it only matches the characters instead of the whole filename.
Here is the online tool: https://www.regextester.com/
Tried with remove .
My guess is that you might be trying to design an expression similar to:
^(?!.*\d).*$
The expression is explained on the top right panel of this demo, if you wish to explore/simplify/modify it, and in this link, you can watch how it would match against some sample inputs step by step, if you like.
try:
/^(\D*)$/gm
Hope it helps
François
You can simply use
^[^\d]+$
Demo
The following regular expression should do the job. Since you've requested to only capture the filename of a file not containing numbers (this means only the filename and not the extension of the file).
^([a-zA-Z]+)(?=\.[a-zA-Z])
You can test the above regular expression here:
https://rubular.com/r/NMcMicEmLUKNTB
Related
I have the following text, for example:
nino&searchPhrase=jn123456&alphabetical
And I want to extract jn123456.
I've put together the following regex to extract NINOs:
(\bnino?\b.*?|Nino?\b.*?)[a-zA-Z]{2}[0-9]{6}
The problem I have is at the very end of the regex where I'm matching the last alpha character which may or may not be there.
I've tried adding the following at the end of the regex shown above without any luck:
?[a-zA-Z]{1} and
[?a-zA-Z]{1}
Could someone please look at this and let me know where I've gone wrong.
Many thanks and kind regards
Chris
You may use something like this:
^[Nn]ino&?\w*=([a-z]{2}\d{6})
which will capture "jn123456" in the first capturing group.
Demo.
If the character & can be anything else, then you may use . instead.
I am trying to use a regular expression to find a part of a string and select everything up to that string. So for example if my string is this/is/just.some/test.txt/some/other, I want to search for .txt and when I find it select everything before and up to .txt.
After executing the below regex, your answer is in the first capture.
/^(.*?)\.txt/
This matches everything up to ".txt" (without including it):
^.*(?=(\.txt))
You could just do ...
(.*?)\.txt
tested here..
^(.*)text
this worked for me but I was actually trying to get everything after the string too. The first part of the expression should answer your question.
^(.*)text([\s\S]*)$
where ^(.*) takes all before and including the text, while ([\s\S]*)$ takes all after and including the text. Tested it at regexr.com/6cpqg.
References:
above answers among other online sources
https://regexland.com/regex-match-after-character/
Up to and including txt you would need to change your regex like so:
^(.*?\\.txt)
((\n.*){0,3})(.*)\W*\.txt
This will select all the content before the particular word ".txt" including any context in different lines up to 3 lines
It happens all the time, I would need to scan my code for places where I have two or more of the same keywords.
For example $json["VALID"]
So, I would need to find json, and VALID.
Some places in the code may contain:
// a = $json['VALID']; // (note the apostrophes)
(I am using EditPlus which is a great text editor, letting me use regex in my searches)
What would be the string in the regex to find json and VALID (in this example) ?
Thanks in advance!
Use this regex:
\$json\[["']VALID['"]\]
wound find $json<2 character>VALID
\$json.{2}VALID
I have a string that looks like:
this is a string [[and]] it is [[awesome|amazing]]
I have the following regular expression so far:
(?<mygroup>(?<=\[\[).+?(?=\]\]))
I am basically trying to capture everything inside the brackets. However, I need to add another condition that says: If the matched result contains a pipe delimiter then only return the word to the right of the pipe delimiter. If there is no pipe then just return everything inside the brackets.
The parsing result I am looking for given the example above should look like:
and
amazing
Any input is appreciated.
(?<mygroup>(?<=\[\[)([^|\]]*|)?([^|]+?)(?=\]\]))
You could use this regex:
(?<=\[\[[^\]]*?)(?!\w+\|)\w+(?=\]\])
it matches both and and amazing words in your test example. You could check it out, I created a test app on Ideone.
From the regex info page:
The tremendous power and expressivity
of modern regular expressions can
seduce the gullible — or the foolhardy
— into trying to use regexes on every
string‐related task they come across.
My advice: Just grab what is between the brackets and parse it after.
Regular expressions are not the answer to everything. May those who follow after you be spared from deciphering the regex you come up with.
I want to get just the filename using regex, so I've been trying simple things like
([^\.]*)
which of course work only if the filename has one extension. But if it is adfadsfads.blah.txt I just want adfadsfads.blah. How can I do this with regex?
In regards to David's question, 'why would you use regex' for this, the answer is, 'for fun.' In fact, the code I'm using is simple
length_of_ext = File.extname(filename).length
filename = filename[0,(filename.length-length_of_ext)]
but I like to learn regex whenever possible because it always comes up at Geek cocktail parties.
Try this:
(.+?)(\.[^.]*$|$)
This will:
Capture filenames that start with a dot (e.g. .logs is a file named .logs, not a file extension), which is common in Unix.
Gets everything but the last dot: foo.bar.jpeg gets you foo.bar.
Handles files with no dot: secret-letter gets you secret-letter.
Note: as commenter j_random_hacker suggested, this performs as advertised, but you might want to precede things with an anchor for readability purposes.
Everything followed by a dot followed by one or more characters that's not a dot, followed by the end-of-string:
(.+?)\.[^\.]+$
The everything-before-the-last-dot is grouped for easy retrieval.
If you aren't 100% sure every file will have an extension, try:
(.+?)(\.[^\.]+$|$)
how about 2 captures one for the end and one for the filename.
eg.
(.+?)(?:\.[^\.]*$|$)
^(.*)\\(.*)(\..*)$
Gets the Path without the last \
The file without extension
The the extension with a .
Examples:
c:\1\2\3\Books.accdb
(c:\1\2\3)(Books)(.accdb)
Does not support multiple . in file name
Does support . in file path
I realize this question is a bit outdated, however, I had some trouble finding a good source and wound up making the regex myself. To save whoever may find this time,
If you're looking for a ~standalone~ regex
This will match the extension without the dot
\w+(?![\.\w])
This will always match the file name if it has an extention
[\w\. ]+(?=[\.])
Ok, I am not sure why I would use regular expression for this. If I know for example that the string is a full filepath, then I would use another API to get the file name. Regular expressions are very powerfull but at the same time quite complex (you have just proved that by asking how to create such a simple regex). Somebody said: you had a problem that you decided to solve it using regular expressions. Now you have two problems.
Think again. If you are on .NET platform for example, then take a look at System.IO.Path class.
I used this pattern for simple search:
^\s*[^\.\W]+$
for this text:
file.ext
fileext
file.ext.ext
file.ext
fileext
It finds fileext in the second and last lines.
I applied it in a text tree view of a folder (with spaces as indents).
Just the name of the file, without path and suffix.
^.*[\\|\/](.+?)\.[^\.]+$
Try
(?<=[\\\w\d-:]*\\)([\w\d-:]*)(?=\.[\.\w\d-:]*)
Captures just the filename of any kind within an entire filepath. Purposefully excludes the file path and the file extension
Etc:
C:\Log\test\bin\fee105d1-5008-410c-be39-883e5e40a33d.pdf
Doesn't capture (C:\Log\test\bin)
Captures (fee105d1-5008-410c-be39-883e5e40a33d)
Doesn't capture (.pdf)
This RegExp works for me:
(.+(?=\..+$))|(.+[^\.])
Results (bold means match):
test.txt
test 234!.something123
.test
.test.txt
test.test2.txt
.