I am trying to learn regular expressions in vim and have gotten stuck on the following task:
I want to make a substitute which matches all of the following lines - and similar - except for the one containing Look aXXhead. With similar I mean just anything in between the a and the head except XX.
Look ahead
Look aYhead
Look aXhead
Look aXXhead
Look aXXXhead
In case it is relevant I had my highest hopes when trying
:%s/Look a\(XX\)\#!head/*\0
:%s/Look a\(.*\&\(XX\)\#!\)head/*\0
which only matches Look ahead due to the zero width of the \(XX\)\#! I pressume.
Tries like
:%s/Look a\(XX\)\#!.*head/*\0
misses the "triple X" line.
So, how is it done?
P.S. This is my first post on stack exchange so please help and correct me in case there is better ways or places for this question.
Try with this pattern:
/\(Look aXXhead\)\#!\&Look a.*head/
It's made of tho parts:
\(Look aXXhead\)\#!
match at every position in the file BUT not at /Look aXXhead/
Look a.*head
match also /Look aXXhead/
The \& operator tells that each of these parts have to match at the same pos.
how about:
/\vLook a(XXhead)#!.*head
or without very magic:
/Look a\(XXhead\)\#!.*head
Related
I'm trying to see if its possible to extend an existing arbitrary regex by prepending or appending another regex to match within matches.
Take the following example:
The original regex is cat|car|bat so matching output is
cat
car
bat
I want to add to this regex and output only matches that start with 'ca',
cat
car
I specifically don't want to interpret a whole regex, which could be quite a long operation and then change its internal content to match produce the output as in:
^ca[tr]
or run the original regex and then the second one over the results. I'm taking the original regex as an argument in python but want to 'prefilter' the matches by adding the additional code.
This is probably a slight abuse of regex, but I'm still interested if it's possible. I have tried what I know of subgroups and the following examples but they're not giving me what I need.
Things I've tried:
^ca(cat|car|bat)
(?<=ca(cat|car|bat))
(?<=^ca(cat|car|bat))
It may not be possible but I'm interested in what any regex gurus think. I'm also interested if there is some way of doing this positionally if the length of the initial output is known.
A slightly more realistic example of the inital query might be [a-z]{4} but if I create (?<=^ca([a-z]{4})) it matches against 6 letter strings starting with ca, not 4 letter.
Thanks for any solutions and/or opinions on it.
EDIT: See solution including #Nick's contribution below. The tool I was testing this with (exrex) seems to have a slight bug that, following the examples given, would create matches 6 characters long.
You were not far off with what you tried, only you don't need a lookbehind, but rather a lookahead assertion, and a parenthesis was misplaced. The right thing is: Put the original pattern in parentheses, and prepend (?=ca):
(?=ca)(cat|car|bat)
(?=ca)([a-z]{4})
In the second example (without | alternative), the parentheses around the original pattern wouldn't be required.
Ok, thanks to #Armali I've come to the conclusion that (?=ca)(^[a-z]{4}$) works (see https://regexr.com/3f4vo). However, I'm trying this with the great exrex tool to attempt to produce matching strings, and it's producing matches that are 6 characters long rather than 4. This may be a limitation of exrex rather than the regex, which seems to work in other cases.
See #Nick's comment.
I've also raised an issue on the exrex GitHub for this.
I have the following text, for example:
nino&searchPhrase=jn123456&alphabetical
And I want to extract jn123456.
I've put together the following regex to extract NINOs:
(\bnino?\b.*?|Nino?\b.*?)[a-zA-Z]{2}[0-9]{6}
The problem I have is at the very end of the regex where I'm matching the last alpha character which may or may not be there.
I've tried adding the following at the end of the regex shown above without any luck:
?[a-zA-Z]{1} and
[?a-zA-Z]{1}
Could someone please look at this and let me know where I've gone wrong.
Many thanks and kind regards
Chris
You may use something like this:
^[Nn]ino&?\w*=([a-z]{2}\d{6})
which will capture "jn123456" in the first capturing group.
Demo.
If the character & can be anything else, then you may use . instead.
My college asked my to provide him with a regex that only matches if the test-string endswith
.rar or .part1.rar or part01.rar or part001.rar (and so on).
Should match:
foo.part1.rar
xyz.part01.rar
archive.rar
part3_is_the_best.rar
Should not match:
foo.r61
bar.part03.rar
test.sfv
I immediately came up with the regex \.(part0*1\.)?rar$. But this does match for bar.part03.rar.
Next I tried to add a negative look behind assertion: .*(?<!part\d*)\.(part\0*1\.)?rar$ That didn't work either, because look around assertions need to be fixed width.
Then I tried using a regex-conditional. But that didn't work either.
So my question: Can this even be solved by using pure regex?
An answer should either contain a link to regex101.com providing a working solution, or explain why it can't work by using pure regex.
You could use lookahead to verify the one case that fails your original regex (.rar with .part part that isn't 0*1) is discredited:
^(?!.*\.part0*[^1]\.rar$).*\.(part0*1\.)?rar$
See it in action
This is an old question, but here's another approach:
(?:\.part0*1\.rar|^(?<!\.)\w+\.rar)$
The idea is to match either:
A string that ends with .part0*1.rar (ie foo.part01.rar, foo.part1.rar, bar.part001.rar), OR
A string that ends with .rar and doesn't contain any other dots (.) before that.
Works on all your test cases, plus your extra foo.part19.rar.
https://regex101.com/r/EyHhmo/2
I have this regex for email validation (assume only x#y.com, abc#defghi.org, something#anotherhting.edu are valid)
/^[a-zA-Z0-9]+#[a-zA-Z0-9]\.(com)|(edu)|(org)$/i
But #abc.edu and abc#xyz.eduorg are both valid as to the regex above. Can anyone explain why that is?
My approach:
there should be at least one character or number before #
then there comes #
there should be at least one character or number after # and before .
the string should end with either edu, com, or org.
Try this
/^[a-zA-Z0-9]+#[a-zA-Z0-9]+\.(com|edu|org)$/i
and it should become clear - you need to group those alternatives, otherwise you can match any string that has 'edu' in it, or any string that ends with org. To put it another way, your version matches any of these patterns
^[a-zA-Z0-9]+#[a-zA-Z0-9]\.(com)
(edu)
(org)$
It's worth pointing out that the original poster is using this as a regex learning exercise. This would be a terrible regex for actual production use! It's a thorny problem - see Using a regular expression to validate an email address for a lot more depth.
Your grouping parentheses are incorrect:
/^[a-zA-Z0-9]+#[a-zA-Z0-9]+\.(com|edu|org)$/i
Can also just use one case as you're using the i modifier:
/^[a-z0-9]+#[a-z0-9]+\.(com|edu|org)$/i
N.B. you were also missing a + from the second set, I assume this was just a typo...
What you have written is the equivalent of matching something that:
Begins with [a-zA-Z0-9]+#[a-zA-Z0-9].com
contains edu
or ends with org
What you were looking for was:
/^[a-z0-9]+#[a-z0-9]+\.(com|edu|org)$/i
Your regex looks ok.
I guess you are looking using a find function in stead of a match function
Without specifying what you use it is a bit difficult, but in Python you would write
import re
pattern = re.compile ('^[a-zA-Z0-9]+#[a-zA-Z0-9]\.(com)|(edu)|(org)$')
re.match('#abc.edu') # fails, use this to validate an input
re.search('#abc.edu') # matches, finds the edu
Try to use it:
[a-zA-Z0-9]+#[a-zA-Z0-9]+.(com|edu|org)+$
U forget about + modificator if u want to catch any combinations of (com|edu|org)
Upd: as i see second [a-zA-Z0-9] u missed + too
I am looking to find anything that matches this pattern, the beginning word will be:
organism aogikgoi egopetkgeopt foprkgeroptk 13
So anything that starts with organism needs to be found using regex.
^organism will match anything starting with "organism".
^organism(.*) will also capture everything that follows, into the variable that contains the first match (which varies according to language -- in Perl it's $1).
Also just wanna add for others newbies like me and their various circumstances, you can do it in various ways depending on your text and what you are tryna do.
Like here's an Example where I wanna delete everything after ?spam so I could use .?spm.+ or .?spm.+ or any other ways as long you are creative about it lol.
This might come in handy, here's a Link | Link where you can find some basic necessary regex and their meanings.