I have a requirement to search through a directory to identify specific files with a string contained in the file name. But I want to exclude part loaded files with a ".filepart" extension.
This must be done through Regex due to tool limitations.
The file names can be in multiple formats, and we must identify them from the "file identifier" string that we pass into the Regex.
I have read some very good articles within SO and other websites but I am struggling to nail down the correct syntax.
I have saved a page on regex101.com to provide a more detailed explanation of what I am trying to achieve. The "FILETYPE" can be considered the string we pass into the Regex.
https://regex101.com/r/zTrbyX/4
Thanks,
K
Your original regex is:
.*FILETYPE.*\.[[:alnum:]]*(?!filepart)
will give the same result as:
.*FILETYPE.*
Instead you could use the following regex (similar to CAustin solution in comments):
.*FILETYPE.*(?<!filepart)$
This will match every line starting with .*FILETYPE.* and not ending with filetpart. Here $ denotes the end of the line. In regex101.com you need to activate flag m for $ to be recognized as EOL.
I have two files in a directory. FileAbc_1.xml and FileAbc.xml. I want write a regex that only select FileAbc_1.xml.
My regex is : FileAbc.*.xml
It is picking up both file names but I only want FileAbc_1.xml. Any help would great favor.
This will work for you
FileAbc_[0-9]+.xml
That should just be: FileAbc_\d\.xml
(assuming there's never more than one digit after the underscore)
You can go with this for anything that will start with FileAbc and end with XML FileAbc.+\.xml.
If I have a list of file names in an XML and want to remove all instances where the file name doesn't have a file extension, how can I do this using regular expressions? I need to do the replace in TextWrangler and have no other option unfortunately.
For example, if I have such a list in an XML as:
<name>AAA_A026C032_150522_R4RO.mov</name>
<name>BBB_A016D032_150809_R4RO.aiff</name>
<name>CCC_A026C038_151010_R4RO</name>
<name>DDGS_A006C132_150409_R4RO.mp3</name>
<name>EFFD_B026C001_150607_R4RO</name>
<name>FGHG_A026C032_141215_R4RO.cine</name>
Have can the files without the file extension be targeted using regular expressions? I would like to replace these (clear them) in the output document.
Thanks in advance,
Matt
'(?!>\w+\.[a-zA-Z0-9]+)>(\w+)'
this pattern gets the name of the files without extensions as its first capturing group. I dont know how to use TextWrangler but I assume that with filename string, you can probably figure it out?
I have a bunch of files, some examples are as follows:
/foo1/foo2/bar1.bar2.bar3.answer.jar
/foo1/bar1.bar2.answer.jar
/foo1/foo2/answer.jar
and for all of the above I would like a regex that matches 'answer'. In other words, I'm looking to get an alias for the file that is the portion of the file name after the last dot (or the file name itself if there are no dots) with the file extension (.jar can be guaranteed here to make it simpler) stripped off.
I know I can do this with a more simple regex to split the value up by dots and then get the second last one, but in this case I'm building a back-end thing that will ideally take a regex that is defined in a configuration definition for the given file type, and spit out the alias, which might be different for other file types.
Yep, I'm over-engineering. :)
Any ideas?
Following regex should work for you:
[^/.]+(?=\.jar$)
If using Javascript or a similar flavor where / is regex delimiter then you need to escape / like this:
[^\/.]+(?=\.jar$)
You can use the following regexp: (assuming that the answer part doesn't contain . or /)
[/\.]([^/\.]+)\.jar
The first capturing group is the part what you want to.
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
.