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Check if word alternates consonant and vowel [closed]

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So I need to check if a word is a pattern of alternating vowel and cosonant (Or consonant and vowel) in Java.
I want to make it a regex but I just came with this incomplete regex expression:
[aeiouAEIOI][^aeiouAEIOI]
Any ideas?
Thanks :)
Update: It's not regex restricted, so it can be an option if anyone has any ideas

One way is using a lookahead to check if neither two vowels nor two consonants next to each other.
(?i)^(?!.*?(?:[aeiou]{2}|[^aeiou]{2}))[a-z]+$
See this demo at regex101 (used i flag for caseless matching, the \n in demo is for staying in line)
Update: Thank you for the comment #Thefourthbird. For matching at least two characters you will need to change the last quantifier: Use [a-z]{2,} (two or more) instead of [a-z]+ (one or more). For only matching an even amount of characters (2,4,6,8...), change this part to: (?:[a-z]{2})+
FYI: If you use this with matches you can drop the ^ start and $ end anchor (see this Java demo).

Java regular expression with hyphen is not count in length [closed]

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How do I can make regular expression, that accept all alphanumeric, except ,(coma).I tried the following expression [^,]{0,10}.but the hyphen is not counting in length.. Pls suggest me. My input is "12345-7890" and "1a2s2-6s7a"

This regex will exclude any pattern that contains a comma.
^((?!\-)[^\,]{0,10}|(?=\-)[^\,]{0,9})$
The regex will detect any hyphen and set the maximum length to 9 (if not contains hyphen) or 10 otherwise. So the hyphen still counts to length, which is adjusted dynamically. This only works with one hyphen.

Regex to replace function foo(a,b) to function foo(b,a) [closed]

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I have function call like BeanUtils.copyProperties(source, destination);
I want to change it to BeanUtils.copyProperties(destination, source); in many places. How to do it using Regex? What is the regex command to do this?
I'm using eclipse to do find and replace.

Search for (with regex setting turned on)
BeanUtils\.copyProperties\s*\(\s*([\w\_]+)\s*\,\s*([\w\_]+)\s*\)\s*\;
And replace with:
BeanUtils.copyProperties($2, $1);
First escape all literal characters with backslash \
Wherever a space can be found when writing code, match it with 0 or more spaces. That by using \s* Could use [ ]* but \s might be sufficient in this case.
Then add captures for the source and destination by adding them in brackets. Or use [\w\_]+ to match other variable names. With a + to mean at least 1 char. NB: if your variable have any other non-alphanumeric chars, add them to the [...] list.
Finally in the replace, switch the captures.

matching patterns with an optional string [closed]

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In a Java app, I use this regex: (\w+)_\d to match patterns of this form:
apples_1
oranges_2
and then I use the first capturing group value (apples, oranges).
However, I now have a new request to also match these strings:
applesdrp_1
orangesdrp_2
where 'drp' is a fixed 3 character string, and the same values as before need to be captured: apples, oranges
So for example, if I use this regex: (\w+)(?:drp)?_\d
it will do the work on apples_1, but not for applesdrp_1.
Is there a way to do that with a regex?

You can use a non-greedy quantifier:
(\w+?)(?:drp)?_\d
In this way \w+? will take characters until it find "drp_N" or "_N" (where N is a digit).
If you use a greedy quantifier, \w+ takes all possible character (including the underscore and the digit since they are included in \w) and then gives back characters one by one until (?:drp)?_\d succeeds. But since (?:drp)? is optional, the regex engine stops to backrack when it find _N.

Yes, you can - one way would be using a negative lookbehind, to make sure, that the drp is forced outside the group, if it is present
(\w+)(?<!drp)(?:drp)?_\d+
See https://regex101.com/r/jJ1rM4/3 for a demo

How this regular expression do this? [closed]

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I want to use below expression in my program but i don't know what do this regular expression!
please help me.
"(?=(?!^)[,;.:])|(?<=[,;.:])"
in the above expression (?=(?!^)[,;.:]) find any character set that end with [.;,:] or no? what do this (?!^) in this expression?
and this expression find any character set that end with [,;.:] or no?
please help me.

The expression matches 0-length strings that satisfy one of these two conditions:
Ahead of it is one of ,;.:, but not for 0-length strings just before the beginning of the subject string (position 0). So the subject string "." has no match at position 0, only at position 1 because of the following alternative. This is done with positive lookahead (?=) and negative lookahead (?!).
Behind it is one of ,;.:. This is done with positive lookbehind (?<=).
For instance for "aaa,1", you have two matches: at position three (after the last a, because it's followed by ,) and at position 4 (because it's preceded by ,).