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Add Dash to Java Regex

I am trying to modify an existing Regex expression being pulled in from a properties file from a Java program that someone else built.
The current Regex expression used to match an email address is -
RR.emailRegex=^[a-zA-Z0-9_\\.]+#[a-zA-Z0-9_]+\\.[a-zA-Z0-9_]+$
That matches email addresses such as abc.xyz#example.com, but now some email addresses have dashes in them such as abc-def.xyz#example.com and those are failing the Regex pattern match.
What would my new Regex expression be to add the dash to that regular expression match or is there a better way to represent that?

Basing on the regex you are using, you can add the dash into your character class:
RR.emailRegex=^[a-zA-Z0-9_\\.]+#[a-zA-Z0-9_]+\\.[a-zA-Z0-9_]+$
add
RR.emailRegex=^[a-zA-Z0-9_\\.-]+#[a-zA-Z0-9_-]+\\.[a-zA-Z0-9_-]+$
Btw, you can shorten your regex like this:
RR.emailRegex=^[\\w.-]+#[\\w-]+\\.[\\w-]+$
Anyway, I would use Apache EmailValidator instead like this:
if (EmailValidator.getInstance().isValid(email)) ....

Meaning of - inside a character class is different than used elsewhere. Inside character class - denotes range. e.g. 0-9. If you want to include -, write it in beginning or ending of character class like [-0-9] or [0-9-].
You also don't need to escape . inside character class because it is treated as . literally inside character class.
Your regex can be simplified further. \w denotes [A-Za-z0-9_]. So you can use
^[-\w.]+#[\w]+\.[\w]+$
In Java, this can be written as
^[-\\w.]+#[\\w]+\\.[\\w]+$

^[a-zA-Z0-9_\\.\\-]+#[a-zA-Z0-9_]+\\.[a-zA-Z0-9_]+$
Should solve your problem. In regex you need to escape anything that has meaning in the Regex engine (eg. -, ?, *, etc.).

The correct Regex fix is below.
OLD Regex Expression
^[a-zA-Z0-9_\\.]+#[a-zA-Z0-9_]+\\.[a-zA-Z0-9_]+$
NEW Regex Expression
^[a-zA-Z0-9_.+-]+#[a-zA-Z0-9-]+\.[a-zA-Z0-9-.]+$

Actually I read this post it covers all special cases, so the best one that's work correctly with java is
String pattern ="(?:[a-zA-Z0-9!#$%&'*+/=?^_`{|}~-]+(?:\\.[a-zA-Z0-9!#$%&'*+/=?^_`{|}~-]+)*|\"(?:[\\x01-\\x08\\x0b\\x0c\\x0e-\\x1f\\x21\\x23-\\x5b\\x5d-\\x7f]|\\\\[\\x01-\\x09\\x0b\\x0c\\x0e-\\x7f])*\")#(?:(?:[a-zA-Z0-9](?:[a-zA-Z0-9-]*[a-zA-Z0-9])?\\.)+[a-zA-Z0-9](?:[a-zA-Z0-9-]*[a-zA-Z0-9])?|\\[(?:(?:(2(5[0-5]|[0-4][0-9])|1[0-9][0-9]|[1-9]?[0-9]))\\.){3}(?:(2(5[0-5]|[0-4][0-9])|1[0-9][0-9]|[1-9]?[0-9])|[a-zA-Z0-9-]*[a-zA-Z0-9]:(?:[\\x01-\\x08\\x0b\\x0c\\x0e-\\x1f\\x21-\\x5a\\x53-\\x7f]|\\\\[\\x01-\\x09\\x0b\\x0c\\x0e-\\x7f])+)\\])";

Validate string has no illegal characters

Im trying to validate a string that only allows letters, numbers and these characters :
!"#$%&'()*+,-./:;<=>?#[\]^_`{|}~
I tried doing this but its not working and allowing me to enter characters not in the regex. Im still pretty new to java and something similar was working in javascript but I cant figure out whats going on here. I think its running as if it cant find any of the characters mentioned then it will return four.
Pattern allowedCharacters = Pattern.compile("[A-Za-z0-9!\"#$%&'()*+,.\\/:;<=>?#[\\]^_`{|}~-]+$]");
if (!allowedCharacters.matcher(pw).find()){
return 4;
}
Any help is appreciated. Thanks
EDIT:
I also tried:
if (pw.matches("^[A-Za-z0-9!\"#$%&'()*+,.\\/:;<=>?#[\\]^_`{|}~-]+$]")){
return 4;
}
and
if (!pw.matches("[A-Za-z0-9!\"#$%&'()*+,.\\/:;<=>?#[\\]^_`{|}~-]+$]")){
return 4;
}

matcher.find() checks if string contains substring that matches regex, so with
!matcher.find() you are checking if there is no match of regex in tested string.
Consider using using matcher.matches() to check if entire string is matched by regex. In this case you will have to add quantifiers like *, + or {n,m} to character class to decide about passwords length. Otherwise it will only single character passwords.
Here is demo of how your code can look like
// here you place quantifier
// ↓
if (pw.matches("[A-Za-z0-9!\"#$%&'()*+,.\\/:;<=>?#[\\]^_`{|}~-]+$]+")){
System.out.println("password contains only valid characters");
} else {
System.out.println("invalid characters in password");
}
Update:
in your regex you are not escaping [ which makes [\]^_`{|}~-] separate character class which will be added to outer character class. This character class will not include \ or [. If you are really interested in accepting only alphanumeric characters and !"#$%&'()*+,-./:;<=>?#[]^_`{|}~ then consider using
"[\\w\\Q!\"#$%&'()*+,-./:;<=>?#[\\]^_`{|}~\\E]+"
as regex.
\\w represents [a-zA-Z0-9_]
and \Q and \E is quote, which is mechanism to escape metacharacters, even in character class.

It's because you're using find() and not matches(). That said, I'd try the opposite, doing find on [^<legal chars>] (note the caret) to match an illegal characters. It's faster because it'll fail as soon as it hits something illegal. Also, start with the simple legal characters, then move up from there. Regular expressions can get hard to read, and adding one char at a time that has special meaning is easier than adding them all at once.

Using other answers from this question, I found this to work for me. Nothing needs to be escaped between the \Q and \E. They do that for you.
Pattern whitelist = Pattern.compile("^[\\w\\s\\Q!\"#$%&'()*+,-.\\/:;<=>?#[]^_`{|}~\\E]+$");
if (!whitelist.matcher(pw).matches()) {
// error
}

Java RegEx pattern is invalid when trying to exclude commas

I'm building a function to validate usernames, and in this case I want to accept alphabetic characters only. I'm matching the provided user input against this regex:
[1-9!##$%&*()_+=|<>?{}\\[\\]~-,]
This is the method that makes use of the regex:
public static String purgeInvalidLogin(String failedLogin, String pattern) {
Pattern special = Pattern.compile (pattern);
String purgedLogin = failedLogin.replaceAll(special.pattern(), ""); // remove any special characters before moving on
purgedLogin = StringUtils.deleteWhitespace(purgedLogin);
return purgedLogin;
}
However when trying to run this I get this message:
Illegal character range near index 25 [!##$%&*()_+=|<>?{}[]~-,] ^
which only happened once I added the comma. I've also tried the expression [!##$%&*()_+=|<>?{}[]~-\,] (escaping the comma) to no avail. I'm wondering how I can use the regex properly to exclude commas making use of my method above.
Thanks in advance.

Escape the hyphen just before it. It is interpreted as defining a range of characters, as soon as you add another character (the comma) after it.
[1-9!##$%&*()_+=|<>?{}\\[\\]~\\-,]

You want to accept only alpha chars and you are doing this by listing every possible illegal character. I think you have got this backwards and it would better to look for what you do want (which would be a much shorter regex) and flag non matches.

Splitting a string in java on more than one symbol

I want to split a string when following of the symbols encounter "+,-,*,/,="
I am using split function but this function can take only one argument.Moreover it is not working on "+".
I am using following code:-
Stringname.split("Symbol");
Thanks.

String.split takes a regular expression as argument.
This means you can alternate whatever symbol or text abstraction in one parameter in order to split your String.
See documentation here.
Here's an example in your case:
String toSplit = "a+b-c*d/e=f";
String[] splitted = toSplit.split("[-+*/=]");
for (String split: splitted) {
System.out.println(split);
}
Output:
a
b
c
d
e
f
Notes:
Reserved characters for Patterns must be double-escaped with \\. Edit: Not needed here.
The [] brackets in the pattern indicate a character class.
More on Patterns here.

You can use a regular expression:
String[] tokens = input.split("[+*/=-]");
Note: - should be placed in first or last position to make sure it is not considered as a range separator.

You need Regular Expression. Addionaly you need the regex OR operator:
String[]tokens = Stringname.split("\\+|\\-|\\*|\\/|\\=");

For that, you need to use an appropriate regex statement. Most of the symbols you listed are reserved in regex, so you'll have to escape them with \.
A very baseline expression would be \+|\-|\\|\*|\=. Relatively easy to understand, each symbol you want is escaped with \, and each symbol is separated by the | (or) symbol. If, for example, you wanted to add ^ as well, all you would need to do is append |\^ to that statement.
For testing and quick expressions, I like to use www.regexpal.com

How to replace brackets in strings

I have a list of strings that contains tokens.
Token is:
{ARG:token_name}.
I also have hash map of tokens, where key is the token and value is the value I want to substitute the token with.
When I use "replaceAll" method I get error:
java.util.regex.PatternSyntaxException: Illegal repetition
My code is something like this:
myStr.replaceAll(valueFromHashMap , "X");
and valueFromHashMap contains { and }.
I get this hashmap as a parameter.

String.replaceAll() works on regexps. {n,m} is usually repetition in regexps.
Try to use \\{ and \\} if you want to match literal brackets.
So replacing all opening brackets by X works that way:
myString.replaceAll("\\{", "X");
See here to read about regular expressions (regexps) and why { and } are special characters that have to be escaped when using regexps.

As others already said, { is a special character used in the pattern (} too).
You have to escape it to avoid any confusion.
Escaping those manually can be dangerous (you might omit one and make your pattern go completely wrong) and tedious (if you have a lot of special characters).
The best way to deal with this is to use Pattern.quote()
Related issues:
How to escape a square bracket for Pattern compilation
How to escape text for regular expression in Java
Resources:
Oracle.com - JavaSE tutorial - Regular Expressions

replaceAll() takes a regular expression as a parameter, and { is a special character in regular expressions. In order for the regex to treat it as a regular character, it must be escaped by a \, which must be escaped again by another \ in order for Java to accept it. So you must use \\{.

You can remove the curly brackets with .replaceAll() in a line with square brackets
String newString = originalString.replaceAll("[{}]", "X")
eg: newString = "ARG:token_name"
if you want to further separate newString to key and value, you can use .split()
String[] arrayString = newString.split(":")
With arrayString, you can use it for your HashMap with .put(), arrayString[0] and arrayString[1]