We Keep Coding

Java regex - Replace "x^2" with "x²" but NOT "x^27" with "x²7"

I got a string of an equation where I want to replace all occurrences of the scheme "x^2" with "x²".
My code:
String equation = "x^2";
equation = equation.replace("^2", "\u00B2"); // u00b2 is unicode for '²'
System.out.println(equation);
This works for "x^2" but for example "x^25" I'm getting the string "x²5", but in such a case I want it to stay the same "x^25".
Another example:
"x^2 + 6x" -> "x² + 6x" // ... x squared
"x^28 + 6x" -> "x^28 + 6x" // ... x to the power of 28
Thank you!
EDIT:
The solution from "Mshnik" works perfectly, even with a custom character like "y^2" instead of "x^2", thanks!

Here's a regex that will match 2 in x^2, the 2 in a^2+... but not the 2 in x^20:
(?<=\w)\^2(?![0-9.])
Specifically:
(?<= <EXP>) is a positive lookbehind on <EXP>, More explanation here
\w matches any alphabetic character, upper or lower case.
\^ matches the ^ character literally
2 matches the 2 character literally
(?! <EXP>) is a negative lookahead on <EXP> More explanation here.
[0-9.] matches all numbers and decimals, like 58 and 3.14.
Thus all together it matches the 2 that is preceded by x^ and isn't followed by a digit.
With that, you can use java's Pattern class to find and rebuild a string with the new ². More on that here
Note that in order to get a backslash into a java regex, you need the literal backslash character represented by \\. Thus the final result looks like (?<=\\w)\\^2(?![0-9.]).

Regex to match only one even decimal digit

I am building on the idea of [Jonah][1]: Matching to one decimal place in Java using regex
^([1-9]\d*|0)(\.\d)?$
This regex is the one I want. But my additional requirement is that the decimal digit must be an even number such as 0.2, 10.4, 100.6, 11.8.
I am thinking of splitting the string at '.' and checking if decimal digit % 2 = 0. How can I achieve this in regex?

Instead of using the /d for a decimal digit, you can just explicitly require an even number, like this:
^([1-9]\d*|0)(\.[24680])?$

Regular expression to match until last 3 characters before a comma

Maybe this is asked somewhere but certainly I couldn't find the answer I want so:
I'm having difficulties to match specific characters in a string:
"88551554,86546546,51516565"
The digits I want to match are the X's in the following :
"XXXXX554,XXXXX546,XXXXX565"
Right now I'm only able to find out the last 3 digits before each comma :
\d{3}(?=,)
And since the length of the numbers are dynamic, it seems not possible to specify the number of digits before the 3 digits.
Anyone can help?
Thanks in advance!

You can use this lookahead regex:
(\d+)(?=\d{3}(?:,|$))
RegEx Demo
This will match and group 1 or more digits that must be followed by 3 digits and a comma or end of input. Check MATCH INFORMATION in the demo link for captured groups.
Update: To replace all those matched digits by X use:
str = str.replaceAll("\\d(?=\\d*\\d{3}(?:,|$))", "X");
RegEx Demo2

To match it use:
\d+(?=\d{3})
This regex does:
\d+... Match a digit (0-9) between one and unlimmited times.
(?=\d{3}) ... Match a digit (0-9) exactly three times inside an positive lookahead.

regex to match a recurring pattern

I am trying to write a regex for java that will match the following string:
number,number,number (it could be this simple or it could have a variable number of numbers, but each number has to have a comma after it there will not be any white space though)
here was my attempt:
[[0-9],[0-9]]+
but it seems to match anything with a number in it

You could try something along the lines of ([0-9]+,)*[0-9]+
This will match:
Only one number, e.g.: 7
Two numbers, e.g.: 7,52
Three numbers, e.g.: 7,52,999
etc.
This will not match:
Things with spaces, e.g.: 7, 52
A list ending with a comma, e.g.: 7, 52,
Many other things out of the scope of this problem.

I think this would work
\d+,(\d+,)+
Note that as you want, that will only capture number followed by a comma

I guess you are starting with a String. Why don't you just use String.split(",") ?

^ means the start of a string and $ means the end. If you don't use those, you could match something in the middle (b matched "abc").
The + works on the element before it. b is an element, [0-9] is an element, and so are groups (things wrapped in parenthesis).
So, the regex you want matches:
The start of the string ^
a number [0-9]
any amount of comas flowed by numbers (,[0-9])+
the end of the string $
or, ^[0-9](,[0-9])+$

Try regex as [\d,]* string representation as [\\d,]* e.g. below:
Pattern p4 = Pattern.compile("[\\d,]*");
Matcher m4 = p4.matcher("12,1212,1212ad,v");
System.out.println(m4.find()); //prints true
System.out.println(m4.group());//prints 12,1212,1212
If you want to match minimum one comma (,) and two numbers e.g. 12,1212 then you may want to use regex as (\d+,)+\d+ with string representation as \\d+,)+\\d+. This regex matches a a region with a number minimum one digit followed by one comma(,) followed by minimum one digit number.

What does this syntax for regular expressions mean?

Can someone tell me what the regular expression in the following Java code snippet means:
String someString = …;
someString.matches("^\\d{5}-\\d{4}$");

This will match 5 decimal numbers at the beginning of the string, followed by a dash, followed by 4 decimal numbers at the end.
^ = Beginning of string
\d{n} = Match n decimal numbers
$ = End of string

From http://docs.oracle.com/javase/6/docs/api/java/util/regex/Pattern.html
five digits, a dash, and then four more digits...nothing else

^ means beginning of line.
\d{5} means five digits.
- literally means "-"
\d{4} means four digits.
$ means end of line.
So it's looking for a sequence of five digits followed by a sequence of four digits, seperated by a dash and that is the only thing on the line.
Example:
12345-6789