First of all, I don't need full e-mail address validation, my given task doesn't require it. I just want to upgrade my current regex code so that it won't match addresses ending with a dot.
My current code: [0-9A-Za-z.]+[#][0-9A-Za-z.]+
It catches both "user#exampe.com", "user#example.com."
I'd like it to catch only from the string that ends without the dot. user#exampe.com
Example string:
dasd.fas#fsaf.dfas.dsa, zghs#gas.gsq, adg32.dsa12#cas, ksak#c.csa., gs32.basaa#scaa.upc.
I'd like to catch the strings marked as code in the example.
Edit: I have only one line with multiple e-mail addresses separated with a , and a space after them.
You might add [0-9A-Za-z]after your regex to end with what you want to match in your character class without the dot followed by a positive lookahead (?=, |$) that asserts what follows is either a comma followed by a whitespace or the end of the string.
[0-9A-Za-z.]+#[0-9A-Za-z.]+[0-9A-Za-z](?=, |$)
Regex Demo
([0-9A-z.]+#(?:\.?[0-9A-z]+)+)(?=,|$)
Try it here
Just slightly modify your pattern: [0-9A-Za-z.]+[#](?:[a-zA-Z]|\.(?=[a-zA-Z]))+.
It uses alternation after # to match one or more: letters OR dot, if it's followed by another letter, thanks to positive lookahead: \.(?=[a-zA-Z]).
Demo
Try this one:
just capture , , $ and group them in non-capturing group except end .
[0-9A-Za-z.]+[#][0-9A-Za-z.]+[0-9A-Za-z](?:(,|$))
demo here
Related
I m trying to improve my regex skills.
I can't manage this exercise.
https://alf.nu/RegexGolf
You have to match words without consecutive identical characters.
To make it clear, we should avoid patterns like abba, or baab, czzc.
The only way I see is to use capture groups:
([a-z])([a-z])\2\1
Then have a negative lookahead:
(?!([a-z])([a-z])\2\1)
But on the site it doesn't work since it doesn't match anything.
Any advice?
Thank you
Use a negative lookahead:
^(?:(.)(?!\1))*$
Explanation:
^ from the start of the input
(?:
(.) match AND capture a single character
(?!\1) then assert that what follows is a different character (not the same)
)* match zero or more such matching characters
$ end of the input
Demo
Another, possibly cleaner, way to do this would be to just have a global negative lookahead at the very start of the pattern:
^(?!.*(.)\1).*$
This would assert at the very beginning that no character is duplicated, anywhere in the string.
^(?!cr|pal|tar)[a-z]{1,4}([a-z])\1[a-z]{0,5}$
This worked for me in the link you gave. I guess we had to match patterns with consecutive letters. But there were some exceptions for which I had to use negative look ahead at the beginning. I have used ([a-z])\1 to match consecutive characters surrounded by possible characters of possible limit. Hope this helps!
Attached the screenshot for reference.
https://i.stack.imgur.com/va1Uq.png
Thanks to Tim Biegeleisen, here is the answer.
^(?!.*(.)(.)\2\1).*$
I need to create regex to find last underscore in string like 012344_2.0224.71_3 or 012354_5.00123.AR_3.335_8
I have wanted find last part with expression [^.]+$ and then find underscore at found element but I can not handle it.
I hope you can help me :)
Just use a negative character class [^_] that will match everything except an underscore (this helps to ensure no other underscores are found afterwards) and end of string $
Pattern would look as such:
(_)[^_]*$
The final underscore _ is in a capturing group, so you are wanting to return the submatch. You would replace the group 1 (your underscore).
See it live: Regex101
Notice the green highlighted portion on Regex101, this is your submatch and is what would be replaced.
The simplest solution I can imagine is using .*\K_, however not all regex flavours support \K.
If not, another idea would be to use _(?=[^_]*$)
You have a demo of the first and second option.
Explanation:
.*\K_: Fetches any character until an underscore. Since the * quantifier is greedy, It will match until the last underscore. Then \K discards the previous match and then we match the underscore.
_(?=[^_]*$): Fetch an underscore preceeded by non-underscore characters until the end of the line
If you want nothing but the "net" (i.e., nothing matched except the last underscore), use positive lookahead to check that no more underscores are in the string:
/_(?=[^_]*$)/gm
Demo
The pattern [^.]+$ matches not a dot 1+ times and then asserts the end of the string. The will give you the matches 71_3 and 335_8
What you want to match is an underscore when there are no more underscores following.
One way to do that is using a negative lookahead (?!.*_) if that is supported which asserts what is at the right does not match any character followed by an underscore
_(?!.*_)
Pattern demo
The pattern:
(test):(thestring)
What I want is full match only if there is just one test: before
test:thestring
But in this case there wouldn't be full match:
test:test:thestring
I've tried qualificator, but it didn't work.
Need help
Try this pattern: ^(?!.*((?(?<=^)|(?<=:))test(?=(:|$))).*(?1)).+$.
The main part is ((?(?<=^)|(?<=:))test(?=(:|$))), which matches test if it's preceeded by colon : or is at the beginning of a line and it's followed by colon : or end of the line.
(?(?<=^)|(?<=:)) this is workaround to (?<=(:|^)), but lookbehinds must have fixed length.
Then we have backreference to first capturing group (?1), to see if there are any other test.
This whole pattern is placed in negative lookahead (?!...), to match everything if it doesn't match pattern explained above (test matched more than one time).
Demo
for this very specific case:
(?<!.)(test:thestring)
Regex101
All it does is search for the string test:thestring and ensures that there are no characters before it. (Use Michał Turczyn's regex for an all purpose search!)
^((?!test:).)*(test:thestring)
See in action
If you want a full match and there should be only one time test: before test:string you might assert the start of the string ^, use a negative lookahead (?:(?!test:).) to match any character if what is on the right side is not test:
Then match test:thestring followed by a negative lookahead (?:(?!test:thestring).)* that matches any character if what is on the right side is not test:thestring and assert the end of the string $
^(?:(?!test:).)*test:thestring(?:(?!test:thestring).)*$
Regex demo
I am trying to extract a string between the two characters # and : for the string:
test23#server:/var/
So, when I try to do something like,
#([^.]*):
or even
\#(\S+):
I get #server:
I just want both # and : removed so I can get just the word server. Please help!
You only need to use a referrer to capturing group you just constructed or you can use \K token. You don't need to escape # character and [^.]* means greedily match everything except a literal dot . which is better to be changed to [^:]+:
#\K[^:]+
or more strictly:
#\K[^:]++(?=:)
Try (?<=#).*(?=:). I'm using positive lookbehind and positive lookahead.
It will match anything starting with (but excluding) # character and ending with (excluding) :
for detail, please see https://www.regular-expressions.info/lookaround.html
see demo at https://regex101.com/r/jIIB9Q/1
For example, for this string I want to match all A and a:
"All the apples make good cake."
Here's what I did: /(.)[^.]*\1*/ig
I started by getting the first character in the group, which can be any character: (.) Then I added [^.]* because I don't want to match any other character that isn't the first one. Finally I added \1* because I wanted to match the first character again. All other similar variations that I've tried don't seem to work.
The regex you are trying to build would capture very first character then any thing up to the same character as much as possible, using a negative lookahead (tempered dot):
(?i)(\w)(?:(?!\1).)*
Capturing group 1 holds the character you need. Try it on a live demo.
If regex engine supports \K match re-setter token then you can append it to the regex above to only match desired part:
(?i)(\w)(?:(?!\1).)*\K