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I am confuse in the logic behind the code (?<=\\D)(?=\\d)|(?<=\\d)(?=\\D)")
it is separating numbers and alphabets like input String abc12dc23 then it is spliting it as output abc 12 dc 23.
I just want the explanation how the above code is working?
This regex:
(?<=\D)(?=\d)|(?<=\d)(?=\D)
matches 2 kinds of patterns, as suggested by the | character:
This pattern:
(?<=\D)(?=\d)
and this pattern:
(?<=\d)(?=\D)
The former looks for a position in the string where there is a non-digit (\D) character before that position and a digit (\d) after it. The latter looks for a position where the reverse happens, a digit before and a non-digit after.
To say this in a more abstract way, the regex is looking for digit-non-digit boundaries.
The split method looks for all occurrences of the pattern and splits the string when it finds one.
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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).
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I am trying to write a boolean function to support double colons :: only for a string. It should reject any string with non-consecutive colon or more than two consecutive colons. The appearance of double colons can be any number. I can write regex which supports double colons but I don't know how to reject so many combinations of non-consecutive and consecutivec colons. Any idea is appreciated!
Valid inputs: Customer::Table, Customer::Table::Sub
Invalid inputs: Customer:Table, Customer::Table:Sub, Customer::::Table
Here is one line option using String#matches:
String input = "Customer::Table::Sub";
if (input.matches("[^:]+(?:::[^:]+)*")) {
System.out.println("MATCH");
}
Demo
Here is an explanation of the regex used:
[^:]+ match one or more non colon characters
(?: start non capturing group
::[^:]+ match :: again followed by one or more non :
)* the group occurring zero or more times
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i have a string like:
String test = "firstName:a,lastName:b,addressOne:line 1,line 2,city:other";
i am trying to do test.replaceAll(",","\",\"").I mean replace , with ",". I want to this only for the whole strings. For eg addressOne is a single string with comma seperated i dont want to replace that. Is there any regex or someother way i can do this?
i should get a string like
"firstName":"a","lastName":"b","addressOne":"line 1 , line 2","city":"other"
after replace , but i am getting
"firstName":"a","lastName":"b","addressOne":"line 1 "," line 2","city":"other".
You can split your string with commas that are followed by a word which has been followed by :. And for this aim you can use a positive look-ahead and for matching the leading word use a negated character class [^:,]+ which will match any string except : and ,:
test.split(",(?=[^:,]+:)")
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I need to take out all the characters of a string that are not numbers.
You could use a Character Class Intersection like [\p{Punct}\p{Lower}\p{Upper}&&[^.]]
But why not just use
[^\d.]+
As Java String "[^\\d.]+"
This would match one or more characters, that are not \d a digit or the . period.
I'd suggest using \\d+ then (it's consecutive digits), and a capture group. Something like
String str = "";
str = str.replaceAll("(\\d+\\.\\d+)", "$1");
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So I have to use Regex for the name of a character, where the name has to start with a capital and can only exist out of letters, apostrophe's and spaces.
And I dont know how to start, could anyone help me with this?
I suppose your character name is stored in a String variable. For that you can use the String.matches() function, which accepts a regex String parameter.
To create the required regex you will have to combine the folowing:
[A-Z] for the capital letter
[a-z' ]+ for the remaining characters.
Note that when using those in Java you'll need to add some escape characters
You can experiment with regular expressions here: http://www.regexr.com