I haven't found what I am looking for, but I am trying to match the last digit of strings. For example, I want to pull out only those that end in the number 9 from the data below. How can this be done in regex? I'm just beginning to play around with it, so any help is appreciated. Thanks!
abc12997654
efg12456789
dkj98765433
kij12444449
yur123ert97
output:
efg12456789
kij12444449
One possible way to do that (of course there are alternatives) is as follows:
It matches the words only ending with a digit 9.
/(\w+9\b)/g
(.*9$) is a simpler regex that will give you your output!
Regex Demo
you can use word boundaries to assert a correct match occurs.
regex: \b\w+9\b
https://regex101.com/r/hAA0pC/1
\b\w+9\b
\b matches a non word character, in this case start and end of the string.
\w matches a alpha-numeric or _ character, + grabs one or more matching characters.
Related
just starting out with regex and have hit a stumbling block. Hoping someone might be able to explain the workaround.
Trying to carry out a multi-line search. I wish to use "*" as the 'flag', so to speak: if a line contains an asterisk it should match. The digits at the start of the line should be output, so should the word "Match" in the linked example, excluding the asterisk itself.
I assume my use of "|" is dividing the regex into two conditions, when it actually needs to satisfy both to match.
https://regex101.com/r/Pu56bi/2
(?m)(^\d+)|(?<=\*).*$
Any help kindly appreciated.
You could use a pos. lookahead as in
^(?=.*?\*)(\d+).+?(Match)$
See your modified example on regex101.com.
If Match is always at the end of the string, you could match the digits at the start of the string, then match an * and Match at the end of the string.
Use a word boundary \b to prevent the word of digits being part of a longer word.
^(\d+)\b.*\*.*\b(Match)$
Regex demo
If there can be test after the word Match you can assert * using a positive lookahead.
^(?=.*\*)(\d+)\b.*\b(Match)\b.*$
Regex demo
I need to get a regex that will find a match of a single lower case a-z character followed by 5 numbers that is either:
at the start of a line
at the end of a line
surrounded by () or []
surrounded by whitespace
So the following results are expected:
a12345 MATCH
(a12345) MATCH
[a12345] MATCH
text a12345 MATCH
aa12345 NO MATCH
At the moment I have this (?<=[])]*)[a-z]{1}[0-9]{5}(?=[])]*) but it is not working for all scenarios, for example it sees aa12345 and a12345a as being matches when I don't want them to.
Can anyone help?
EDIT:
Apologies I should have mentioned this is for .NET c#
First of all your should mention programming language.
Following solution is for PCRE.
Regex: ((?<=[\[( ])|^)[a-z]\d{5}((?=[\]\) ])|$)
Explanation:
((?<=[\[( ])|^) checks for preceding brackets, whitespaces OR beginning.
[a-z]\d{5} checks for alphabet followed by 5 digits.
((?=[\]\) ])|$) checks for succeeding brackets, whitespaces OR end of line.
Regex101 Demo
Does this work:
(\[[a-z]\d{5}\])|(\([a-z]\d{5}\))|(\b[a-z]\d{5}\b)
Using Regular Expression,
from any line of input that has at least one word repeated two or more times.
Here is how far i got.
/(\b\w+\b).*\1
but it is wrong because it only checks for single char, not one word.
input: i might be ill
output: < i might be i>ll
<> marks the matched part.
so, i try to do (\b\w+\b)(\b\w+\b)*\1
but it is not working totally.
Can someone give help?
Thanks.
this should work
(\b\w+\b).*\b\1\b
greedy algorithm will ensure longest match. If you want second instance to be a separate word you have to add the boundaries there as well. So it's the same as
\b(\w+)\b.*\b\1\b
Positive lookahead is not a must here:
/\b([A-Za-z]+)\b[\s\S]*\b\1\b/g
EXPLANATION
\b([A-Za-z]+)\b # match any word
[\s\S]* # match any character (newline included) zero or more times
\b\1\b # word repeated
REGEX 101 DEMO
To check for repeated words you can use positive lookahead like this.
Regex: (\b[A-Za-z]+\b)(?=.*\b\1\b)
Explanation:
(\b[A-Za-z]+\b) will capture any word.
(?=.*\b\1\b) will lookahead if the word captured by group is present or not. If yes then a match is found.
Note:- This will produce repeated results because the word which is matched once will again be matched when regex pointer captures it as a word.
You will have to use programming to strip off the repeated results.
Regex101 Demo
I'm trying to match the last four characters (alphanumeric) of all words beginning with the sequence &c.
For instance, in the string below, I'd like to match the pieces in bold:
Colour one is &cFF2AC3 and colour two is &c22DE4A.
Can anybody help me with the correct regex expression? I've spent hours on this great resource to no avail.
it looks like hexadecimal numbers, so use this pattern
&c[0-9A-F]{2}\K([0-9A-F]{4})
DEMO
This:
/(?i)\s*&c(?:[a-z0-9]{2})([a-z0-9]{4})\b/
append a g to the end of it if you want it to find all matches in a given text
Try this
/(?:^| )&c\w*(\w{4})\b/
If you want to try it in the regex tester you linked to, make sure to use the g modifier to see all matches.
Explanation: (?:^| ) matches either a space or the start of the string, &c\w* matches the ampersand and the the first however many characters of the word, and then \w{4} captures the last 4 characters. \b on the end asserts a word break (a "non-word" character or the end of the string).
I am pretty weak writing RegExp here is what I'm trying to match:
foo.123
foo.bar.456
anything.here.again.999
In fact, I need to match as many [a-z] founds, seperated by one . character and finally, a number of 3 characters length.
Here is my RegExp: ^([a-z\.]{1,})([0-9]{3})$ the problem is that it does match:
foo.bar111 or even foo..bar.111 is being matched.
Any help would be appreciated.
Thanks
Use the below regular expression. Note the \. outside of the character class:
^([a-z]+\.)+([0-9]{3})$
{1,} can also be replaced with +.
Match start of the string, any number of lower-case letters followed by a dot, and finally any 3 digits followed by end of line:
^([a-z]+\.)+\d\d\d$