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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
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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'm working on a regex for getting a specific number pattern from the URL string.
Requirements: Desire string should start from - or /, followed by a sequence of digits and ending with a / or nothing.
I tried: [-\/](\d+)(\/|$), but for e.g. in www.abc.com/pages/Toms-1777/14623420046 I want /14623420046(i.e. the second occurring digit sequence), but according to my regex, the result will be -1777/. I was trying negative lookbehind but not able to make any progress. I'm new to all this. Please guide.
Test cases: (with matched pattern)
www.abc.com/pages/Essen-Massage-Therapy-LLC/130561253629638
www.abc.com/biz/finn-mccools-santa-monica-2
www.abc.com/summerset.gardens.7
www.abc.com/pages/Toms-1777/14623420046
www.abc.com/pages/The-Clean-Masters/1403753595526512
www.abc.com/24hfsheepsheadbay
www.abc.com/sample2NVCoolSpace
www.abc.com/pages/Jet-Set-3920/542495615847409
www.abc.com/temp.buildings.77
www.abc.com/2423423453534temp/2312312312312312312
www.abc.com/Ptemp-Gtemp-Dtemp-189398324428792/temp
You want that $ in either case. Instead of 'slash OR end', it's more 'optional slash and then a very much not-optional end'. So.. /?$. You don't need to normally escape slashes, especially in java regexes:
Pattern.compile("[-/](\\d+)/?$")
This reads: From minus or slash, a bunch of digits, then 0 or 1 slashes, then the end. Note, use find() and not matches() - matches only works if the entire string matches, which it won't, as the - or / occurs halfway through.
EDIT: Was missing a backslash in the java string.
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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.
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As the title states, i want to know how i can check if a string consists of 2 integers with a blank space in between them in Java.
As an example:
0 2, should return true.
0 abc, should return false.
abcsd, should return false.
And so on...
If it is to any help, I am getting my string from a text file with a buffered reader. Maybe there is a more direct and easier way?
Thank you in advance
You could use string.matches method.
string.matches("\\d+\\s\\d+");
DEMO
\d+ matches one or more digits. So this asserts that the input string must contain a number at the start.
\s matches a space character. So this asserts that the input string must contain a space at the middle.
\d+ matches one or more digits. So this asserts that the input string must contain a number at the end.
Since matches method tries to match the whole input string, you don't need to add start and end anchors to your regex.
Because you haven't posted your own code here, I assume that you haven't made much research into it, have you? First of all, will you use only 1-digit numbers? Here's how you should start: http://docs.oracle.com/javase/tutorial/java/data/converting.html
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I got this regex from a member on this site and decided to play with it till I got what I wanted.
"(\\d+\\.\\d$)", "$10"
It seems to work. My job is to convert any value like 12.3 to 12.30 or 1.0 to 1.00 and 12.33 remains the same.
Does this fit it?
\d would match a single digit
+ is quantifier which matches 1 to many preceding characters..So,\d+ would match 1 to many digits
You can capture values in a group represented as () and then you can refer it back.So when you use this regex (\d)(\d) it captures first digit in group 1 and the second digit in group 2.You can now refer back to these captured values using $1,$2..
One or more numeric digits, followed by a period, followed by a digit and the end of the line
\d matches a single numeric digit, and the + after it means "one or more of what preceded". Then the . is escaped so its just a . , then a single digit again \d, and end of line $.
Note that in java, you need to "double escape" as in \d.