i have regex characters and words as in regex but i want only single member to be selected from this group \d(?i)(R|k|M|E|next|prev){1,2}
valid are - 8RK, 6ME, 9Rnext
invalid are - 8MM,0RR, 9nextnext
Please suggest
As said in the comments, you might want to use lookarounds, namely a neg. lookahead here:
\d(?i)
(?:
(R|k|M|E|next|prev) # capture group 1
(?!\1) # make sure, there's not the same submatch in front
){1,2}
See a demo on regex101.com.
This regex should work ：)
\d\w\w,\s\d\w\w,\s\w.*$
Related
I have a pretty basic regex for a soundcloud link as follows:
/^(https?:\/\/)?((www)?\.)?soundcloud\.com\/[a-zA-Z0-9]*/
It doesn't do much and I would like to block that kind of more complicated links that contain something else (?fbclid=) than only valid soundcloud information:
Invalid:
https://soundcloud.com/beat-one/done-lam/w-avlwV?fbclid=IwAR8TyDI-1ezXHrP1BfaBG1rLP8gTlEQeGJzKO0DGZl7h1i63VWqenV_yDI8
Valid:
https://soundcloud.com/beat-one/done-lam/w-avlwV
How can I do it?
Thanks!
Using a pattern like ((www)?\.)? matches optional www but could also match an optional .
I think you meant to use optional www. as a whole.
For the question mark part, you could match the beginning of the use and then use a negative lookahead (?!.*\?fbclid=) to assert what is on the right does not match ?fbclid=
If that is the case, you could match for example 0+ times a non whitespace char \S*
If you are not using the capturing groups for further processing, you could also make them non capturing (?:
^(https?:\/\/)?(www\.)?soundcloud\.com\/(?!.*\?fbclid=)\S*$
Regex demo
If the link should not contain a question mark, you could simplify the lookahead to (?!.*\?)
Okay, so the task is that there is a string that can either look like post, or post put or even get put post. All of these must be matched. Preferably deviances like [space]post, or get[space] should not be matched.
Currently I came up with this
^(post|put|delete|get)(( )(post|put|delete|get))*$
However I'm not satisfied with it, because I had to specify (post|put|delete|get) twice. It also matches duplications like post post.
I'd like to somehow use a backreference(?) to the first group so that I don't have to specify the same condition twice.
However, backreference \1 would help me only match post post, for example, and that's the opposite of what I want. I'd like to match a word in the first capture group that was NOT previously found in the string.
Is this even possible? I've been looking through SO questions, but my Google-fu is eluding me.
If you are using a PCRE-based regex engine, you may use subroutine calls like (?n) to recurse the subpatterns.
^(post|put|delete|get)( (?!\1)(?1))*$
^^^^
See the regex demo
Expression details:
^ - start of string
(post|put|delete|get) - Group 1 matching one of the alternatives as literal substrings
( (?!\1)(?1))* - zero or more sequences of:
- a space
(?!\1) - a negative lookahead that fails the match if the text after the current location is identical to the one captured into Group 1 due to backreference \1
(?1) - a subroutine call to the first capture group (i.e. it uses the same pattern used in Group 1)
$ - end of string
UPDATE
In order to avoid matching strings like get post post, you need to also add a negative lookahead into Group 1 so that the subroutine call was aware that we do not want to match the same value that was captured into Group 1.
^((post|put|delete|get)(?!.*\2))( (?1))*$
See the regex demo
The difference is that we capture the alternations into Group 2 and add the negative lookahead (?!.*\2) to disallow any occurrences of the word we captured further in the string. The ( (?1))* remains intact: now, the subroutine recurses the whole Capture Group 1 subpattern with the lookahead.
I'm attempting this challenge:
https://regex.alf.nu/4
I want to match all strings that don't contain an ABBA pattern.
Match:
aesthophysiology
amphimictical
baruria
calomorphic
Don't Match
anallagmatic
bassarisk
chorioallantois
coccomyces
abba
Firstly, I have a regex to determine the ABBA pattern.
(\w)(\w)\2\1
Next I want to match strings that don't contain that pattern:
^((?!(\w)(\w)\2\1).)*$
However this matches everything.
If I simplify this by specifying a literal for the negative lookahead:
^((?!agm).)*$
The the regex does not match the string "anallagmatic", which is the desired behaviour.
So it looks like the issue is with me using capturing groups and back-references within the negative lookahead.
^(?!.*(.)(.)\2\1).+$
^^
You can use a lookahead here.See demo.The lookahead you created was correct but you need add .* so that it cannot appear anywhere in the string.
https://regex101.com/r/vV1wW6/39
Your approach will also work if you make the first group non capturing.
^(?:(?!(\w)(\w)\2\1).)*$
^^
See demo.It was not working because \2 \1 were different than what you intended.In your regex they should have been \3 and \2.
https://regex101.com/r/vV1wW6/40
This text
"dhdhd89(dd)"
Matched against this regex
.+?(?:\()
..returns "dhdhd89(".
Why is the start parenthesis included in the capture?
Two different tools, as well as the .NET Regex class, returns the same result. So I gather there is something I don't understand about this.
The way I read my regex is.
Match any character, at least one occurrence. As few as possible.
The matched string should be followed by a start parenthesis, but not to be included in the capture.
I can find workaround, but I still want to know what is going on.
Just turn the non-capturing group to positive lookahead assertion.
.+?(?=\()
.+? non-greedy match of one or more characters followed by an opening parenthesis. Assertions won't match any characters but asserts whether a match is possible or not. But the non-capturing group will do the matching operation.
DEMO
You can just use this negation based regex to capture only text before a literal (:
^([^(]+)
When you use:
.+?(?:\()
Regex engine does match ( after initial text but it just doesn't return that in a captured group to you.
You havn't defined capture groups then I guess you display the whole match (group 0), you can do:
(.+?)(?:\()
and the string you want is in group 1
or use lookahead as #AvinashRaj said.
I'm attempting to capture the 6 digit number in the following:
ObjectID: !nrtdms:0:!session:slonwswtest1:!database:TEST:!folder:ordinary,486150:
I tried the following regex:
\d+(?::$)
attempting to use a non-capturing group to strip the colon out of the returned match, but it returns the colon as in:
486150:
Any ideas what I'm doing wrong?
You want a positive lookahead:
\d+(?=:$)
A non-capturing group is simply a group that cannot be accessed via a backreference; they still are part of the match, nonetheless.
Alternatively, you can use
(\d+):$
and obtain the 1st match group.
You should use a positive lookahead rather than a non-capturing group
\d+(?=:$)
Non-capturing groups are groups that will not create a capture (to be used in backreferences or extracted from the match result). Nonetheless they will match the expression.
What you're looking for is lookahead - to test the expression but exclude it from the match:
\d+(?=:$)
Probably your regex tool is returning the complete match since you don't have any capture group there. Try to enclose the \d+ in a capture group, and find the way to get capture group 1 in your regex tool.
Alternatively, you can also use positive look-ahead:
\d+(?=:$)
And given that you want to capture 6 digits, you can make that explicit:
\d{6}