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Why does this regex not match

I'm building a regex pattern to validate my String inputted.
Here is the limitations that i have to include;
The letters a to z (upper and lowercase) (zero or many times)
o The numbers 0 to 9 (between zero and three times)
o The ampersand (&) character (zero or may times)
o The space character (zero or one time)
Here is the regex I built and tested on
[a-zA-Z&]*|[0-9]{0,3}|[\s]?
String p1 = "[a-zA-Z\\&]*|[0-9]{0,3}|[\\s]?" ;
if (bankName.matches(p1) && bankName.length() >= 8) {
System.out.println("Yes");
}
else{
System.out.println("NO");
}
Here is the entries I'm testing against.
tXiPkaodan57yzrCxYjVT
String bankName = "tXiPkaodan57yzrCxYjVT" ;
On the site i'm testing the regex on is not matching because the numbers ( 5 & 7 ) started and is between the letters but I have included in my regex pattern that it should beable to include any numbers range from 0-9, and 0-3 times
https://www.freeformatter.com/java-regex-tester.html#ad-output
The site i tested it on

The bars ('or') in your regexes have the most precedence, so, that regexp reads:
Match if the input is precisely one of:
Either 0 or more characters that ALL match any letter (either case) or an ampersand,
anywhere between 0 and 3 digits
0 or 1 space.
Your input is none of those things; it is a mix of those things. For example, the input 'a1a' does not match your regexp because, well, step through it: the 'a' forces the regexp matcher to pick the first of your 3 options above, and that's.. it. There's no going back now. Your regexp will match the a, fails to match on the 1, and that's the end of it.
So, how do you fix it? Not by sticking with regexp; that's not good solution for this problem. A regexp that precisely does what you asked for is very convoluted.
Instead, why not just loop through each character, and have 4 counters (spaces, digits, letters, and other stuff). For each character increment the right counter. Then at the end, check that the 'other stuff' counter is 0, digits is 3 or less, and spaces is 1 or less, and then it is valid. Otherwise, it is not.

Java regular expression boundary match?

I found the following question in one Java test suite
Pattern p = Pattern.compile("[wow]*");
Matcher m = p.matcher("wow its cool");
boolean b = false;
while (b = m.find()) {
System.out.print(m.start() + " \"" + m.group() + "\" ");
}
where the output seems to be as follows
0 "wow" 3 "" 4 "" 5 "" 6 "" 7 "" 8 "" 9 "oo" 11 "" 12 ""
Up till the last match it is clear, the pattern [wow]* greedily matches 0 or more 'w' and 'o' characters, while for unmatching characters, including spaces, it results in empty strings. However after matching the last 'l' with 11 "", the following 12 "" seems to be unclear. There is no detailing for this in the test solution, nor I was really able to definitely figure it out from javadoc. My best guess is boundary character, but I would appreciate if someone could provide an explanation

The reason that you see this behavior is that your pattern allows empty matches. In other words, if you pass it an empty string, you would see a single match at position zero:
Pattern p = Pattern.compile("[wow]*"); // One of the two 'w's is redundant, but the engine is OK with it
Matcher m = p.matcher(""); // Passing an empty string results in a valid match that is empty
boolean b = false;
while (b = m.find()) {
System.out.print(m.start() + " \"" + m.group() + "\" ");
}
this would print 0 "" because an empty string is as good a match as any other match for the expression.
Going back to your example, every time the engine discovers a match, including an empty one, it advances past it by a single character. "Advancing by one" means that the engine considers the "tail" of the string at the next position. This includes the time when the regex engine is at position 11, i.e. at the very last character: here, the "tail" consists of an empty string. This is similar to calling "wow its cool".substring(12): you would get an empty string in that case as well.
The engine consider an empty string a valid input, and tries to match it against your expression, as shown in the example above. This produces a match, which your program properly reports.

[wow]* Matches the first wow string. count = 1
Because of the * (zero or more) next to the character class, [wow]* this regex would match an empty string which exists before the character which is not matched by the above pattern. So it matches the boundary or empty space which exists just before to the first space. Count = 2.
its is not matched by the above regex . So it matches the empty string which exists before each character. So count is 2+3=5.
And also the second space is not matched by the above regex. So we get an empty string as match. 5+1=6
c is not matched by the above regex. So it matches the empty space which exists just before to the c 6+1=7
oo is matched by the above regex. [wow]*. So it matches oo and this is considered as 1 match . So we get 7+1=8 as count.
l is not matched. Count = 9
At the last it matches the empty string which exists next to the last character. So now the count is 9+1=10
And finally we all know that the m.start() prints the starting index of the corresponding match.
DEMO

The regex is simply matching the pattern against the input, starting at a given offset.
For the last match, the offset of 12 is at the point after the last character of 'cool' - you might think this is the end of the string and therefore cannot be used for matching purposes - but you'd be wrong. For pattern-matching, this is a perfectly valid starting point.
As you state, your regex expression includes the possibility of zero characters and indeed, this is what happens after the end of the last character, but before the end-of-string marker (usually represented by $ in a regex expression).
To put it another way, without testing past the end of the last character, it would mean no matches would ever occur relating to the end of the string - but there are many regex constructs that match the end of the string (and you've shown one of them here).

Need help for writing regular expression

I am weak in writing regular expressions so I'm going to need some help on the one. I need a regular expression that can validate that a string is an set of alphabets (the alphabets must be unique) delimited by comma.
Only one character and after that a comma
Examples:
A,E,R
R,A
E,R
Thanks

You can use a repeated group to validate it's a comma separated string.
^[AER](?:,[AER])*$
To not have unique characters, you would do something like:
^([AER])(?:,(?!\1)([AER])(?!.*\2))*$

If I understand it correctly, a valid string will be a series (possibly zero long) of two-character patterns, where each pattern is a letter followed by a comma; finally followed at the end by one letter.
Thus:
"^([A-Za-z],)*[A-Za-z]$"
EDIT: Since you've clarified that the letters have to be A, E, or R:
"^([AER],)*[AER]$"

Something like this "^([AER],)*[AER]$"
#Edit: regarding the uniqueness, if you can drop the "last character cannot be a comma" requirement (which can be checked before the regex anyway in constant time) then this should work:
"^(?:([AER],?)(?!.*\\1))*$"
This will match A,E,R, hence you need that check before performing the regex. I do not take responsibility for the performance but since it's only 3 letters anyway...
The above is a java regex obviously, if you want a "pure one" ^(?:([AER],?)(?!.*\1))*$
#Edit2: sorry, missed one thing: this actually requires that check and then you need to add a comma at the end since otherwise it will also match A,E,E. Kind of limited I know.

My own ugly but extensible solution, which will disallow leading and trailing commas, and checks that the characters are unique.
It uses forward-declared backreference: note how the second capturing group is behind the reference made to it (?!.*\2). On the first repetition, since the second capturing group hasn't captured anything, Java treats any attempt to reference text match by second capturing group as failure.
^([AER])(?!.*\1)(?:,(?!.*\2)([AER]))*+$
Demo on regex101 (PCRE flavor has the same behavior for this case)
Demo on RegexPlanet
Test cases:
A,E,R
A,R,E
E,R,A
A
R,E
R
E
A,
A,R,
A,A,R
E,A,E
A,E,E
X,R,E
R,A,E,
,A
AA,R,E

Note: I'm going to answer the original question. That is, I don't care if the elements repeat.
We've had several suggestions for this regex:
^([AER],)*[AER]$
Which does indeed work. However, to match a String, it first has to back up one character because it will find that there is no , at the end. So we switch it for this to increase performance:
^[AER](,[AER])*$
Notice that this will match a correct String the very first time it attempts to. But also note that we don't need to worry about the ( )* backing up at all; it will either match the first time, or it won't match the String at all. So we can further improve performance by using a possessive quantifier:
^[AER](,[AER])*+$
This will take the whole String and attempt to match it. If it fails, then it stops, saving time by not doing useless backing up.
If I were trying to ensure the String had no repeated elements, I would not use regex; it just complicates things. You end up with less-readable code (sadly, most people don't understand regex) and, oftentimes, slower code. So I would build my own validator:
public static boolean isCommaDelimitedSet(String toValidate, HashSet<Character> toMatch) {
for (int index = 0; index < toValidate.length(); index++) {
if (index % 2 == 0) {
if (!toMatch.contains(toValidate.charAt(index))) return false;
} else {
if (toValidate.charAt(index) != ',') return false;
}
}
return true;
}
This assumes that you want to be able to pass in a set of characters that are allowed. If you don't want that and have explicit chars you want to match, change the contents of the if (index % 2 == 0) block to:
char c = toValidate.charAt(index);
if (c == 'A' || c == 'E' || c == 'R' || /* and so on */ ) return false;

How to check a string contains only digits and one occurrence of a decimal point?

My idea is something like this but I dont know the correct code
if (mystring.matches("[0-9.]+")){
//do something here
}else{
//do something here
}
I think I'm almost there. The only problem is multiple decimal points can be present in the string. I did look for this answer but I couldn't find how.

If you want to -> make sure it's a number AND has only one decimal <- try this RegEx instead:
if(mystring.matches("^[0-9]*\\.?[0-9]*$")) {
// Do something
}
else {
// Do something else
}
This RegEx states:
The ^ means the string must start with this.
Followed by none or more digits (The * does this).
Optionally have a single decimal (The ? does this).
Follow by none or more digits (The * does this).
And the $ means it must end with this.
Note that bullet point #2 is to catch someone entering ".02" for example.
If that is not valid make the RegEx: "^[0-9]+\\.?[0-9]*$"
Only difference is a + sign. This will force the decimal to be preceded with a digit: 0.02

If you want to check if a number (positive) has one dot and if you want to use regex, you must escape the dot, because the dot means "any char" :-)
see http://docs.oracle.com/javase/6/docs/api/java/util/regex/Pattern.html
Predefined character classes
. Any character (may or may not match line terminators)
\d A digit: [0-9]
\D A non-digit: [^0-9]
\s A whitespace character: [ \t\n\x0B\f\r]
\S A non-whitespace character: [^\s]
\w A word character: [a-zA-Z_0-9]
\W A non-word character: [^\w]
so you can use something like
System.out.println(s.matches("[0-9]+\\.[0-9]+"));
ps. this will match number such as 01.1 too. I just want to illustrate the \\.

You could use indexOf() and lastIndexOf() :
int first = str.indexOf(".");
if ( (first >= 0) && (first - str.lastIndexOf(".")) == 0) {
// only one decimal point
}
else {
// no decimal point or more than one decimal point
}

I think using regexes complicates the answer. A simpler approach is to use indexOf() and substring():
int index = mystring.indexOf(".");
if(index != -1) {
// Contains a decimal point
if (mystring.substring(index + 1).indexOf(".") == -1) {
// Contains only one decimal points
} else {
// Contains more than one decimal point
}
}
else {
// Contains no decimal points
}

Simplest
Example:
"123.45".split(".").length();

int count=0;
For(int i=0;i<mystring.length();i++){
if(mystring.charAt(i) == '/.') count++;
}
if(count!=1) return false;

Use the below RegEx its solve your proble
allow 2 decimal places ( e.g 0.00 to 9.99)
^[0-9]{1}[.]{1}[0-9]{2}$
This RegEx states:
1. ^ means the string must start with this.
2. [0-9] accept 0 to 9 digit.
3. {1} number length is one.
4. [.] accept next character dot.
5. [0-9] accept 0 to 9 digit.
6. {2} number length is one.
allow 1 decimal places ( e.g 0.0 to 9.9)
^[0-9]{1}[.]{1}[0-9]{1}$
This RegEx states:
1. ^ means the string must start with this.
2. [0-9] accept 0 to 9 digit.
3. {1} number length is one.
4. [.] accept next character dot.
5. [0-9] accept 0 to 9 digit.
6. {1} number length is one.

I create myself to solve exactly question's problem.
I'll share you guys the regex:
^(\d)*(\.)?([0-9]{1})?$
Take a look at this Online Regex to see work properly
Refer documentation if you wish continue to custom the regex
Documentation

Regexp Java for password validation

I'm creating a regexp for password validation to be used in a Java application as a configuration parameter.
The regexp is:
^.*(?=.{8,})(?=..*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[##$%^&+=]).*$
The password policy is:
At least 8 chars
Contains at least one digit
Contains at least one lower alpha char and one upper alpha char
Contains at least one char within a set of special chars (##%$^ etc.)
Does not contain space, tab, etc.
I’m missing just point 5. I'm not able to have the regexp check for space, tab, carriage return, etc.
Could anyone help me?

Try this:
^(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[##$%^&+=])(?=\S+$).{8,}$
Explanation:
^ # start-of-string
(?=.*[0-9]) # a digit must occur at least once
(?=.*[a-z]) # a lower case letter must occur at least once
(?=.*[A-Z]) # an upper case letter must occur at least once
(?=.*[##$%^&+=]) # a special character must occur at least once
(?=\S+$) # no whitespace allowed in the entire string
.{8,} # anything, at least eight places though
$ # end-of-string
It's easy to add, modify or remove individual rules, since every rule is an independent "module".
The (?=.*[xyz]) construct eats the entire string (.*) and backtracks to the first occurrence where [xyz] can match. It succeeds if [xyz] is found, it fails otherwise.
The alternative would be using a reluctant qualifier: (?=.*?[xyz]). For a password check, this will hardly make any difference, for much longer strings it could be the more efficient variant.
The most efficient variant (but hardest to read and maintain, therefore the most error-prone) would be (?=[^xyz]*[xyz]), of course. For a regex of this length and for this purpose, I would dis-recommend doing it that way, as it has no real benefits.

simple example using regex
public class passwordvalidation {
public static void main(String[] args) {
String passwd = "aaZZa44#";
String pattern = "(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[##$%^&+=])(?=\\S+$).{8,}";
System.out.println(passwd.matches(pattern));
}
}
Explanations:
(?=.*[0-9]) a digit must occur at least once
(?=.*[a-z]) a lower case letter must occur at least once
(?=.*[A-Z]) an upper case letter must occur at least once
(?=.*[##$%^&+=]) a special character must occur at least once
(?=\\S+$) no whitespace allowed in the entire string
.{8,} at least 8 characters

All the previously given answers use the same (correct) technique to use a separate lookahead for each requirement. But they contain a couple of inefficiencies and a potentially massive bug, depending on the back end that will actually use the password.
I'll start with the regex from the accepted answer:
^(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[##$%^&+=])(?=\S+$).{8,}$
First of all, since Java supports \A and \z I prefer to use those to make sure the entire string is validated, independently of Pattern.MULTILINE. This doesn't affect performance, but avoids mistakes when regexes are recycled.
\A(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[##$%^&+=])(?=\S+$).{8,}\z
Checking that the password does not contain whitespace and checking its minimum length can be done in a single pass by using the all at once by putting variable quantifier {8,} on the shorthand \S that limits the allowed characters:
\A(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[##$%^&+=])\S{8,}\z
If the provided password does contain a space, all the checks will be done, only to have the final check fail on the space. This can be avoided by replacing all the dots with \S:
\A(?=\S*[0-9])(?=\S*[a-z])(?=\S*[A-Z])(?=\S*[##$%^&+=])\S{8,}\z
The dot should only be used if you really want to allow any character. Otherwise, use a (negated) character class to limit your regex to only those characters that are really permitted. Though it makes little difference in this case, not using the dot when something else is more appropriate is a very good habit. I see far too many cases of catastrophic backtracking because the developer was too lazy to use something more appropriate than the dot.
Since there's a good chance the initial tests will find an appropriate character in the first half of the password, a lazy quantifier can be more efficient:
\A(?=\S*?[0-9])(?=\S*?[a-z])(?=\S*?[A-Z])(?=\S*?[##$%^&+=])\S{8,}\z
But now for the really important issue: none of the answers mentions the fact that the original question seems to be written by somebody who thinks in ASCII. But in Java strings are Unicode. Are non-ASCII characters allowed in passwords? If they are, are only ASCII spaces disallowed, or should all Unicode whitespace be excluded.
By default \s matches only ASCII whitespace, so its inverse \S matches all Unicode characters (whitespace or not) and all non-whitespace ASCII characters. If Unicode characters are allowed but Unicode spaces are not, the UNICODE_CHARACTER_CLASS flag can be specified to make \S exclude Unicode whitespace. If Unicode characters are not allowed, then [\x21-\x7E] can be used instead of \S to match all ASCII characters that are not a space or a control character.
Which brings us to the next potential issue: do we want to allow control characters? The first step in writing a proper regex is to exactly specify what you want to match and what you don't. The only 100% technically correct answer is that the password specification in the question is ambiguous because it does not state whether certain ranges of characters like control characters or non-ASCII characters are permitted or not.

You should not use overly complex Regex (if you can avoid them) because they are
hard to read (at least for everyone but yourself)
hard to extend
hard to debug
Although there might be a small performance overhead in using many small regular expressions, the points above outweight it easily.
I would implement like this:
bool matchesPolicy(pwd) {
if (pwd.length < 8) return false;
if (not pwd =~ /[0-9]/) return false;
if (not pwd =~ /[a-z]/) return false;
if (not pwd =~ /[A-Z]/) return false;
if (not pwd =~ /[%#$^]/) return false;
if (pwd =~ /\s/) return false;
return true;
}

Thanks for all answers, based on all them but extending sphecial characters:
#SuppressWarnings({"regexp", "RegExpUnexpectedAnchor", "RegExpRedundantEscape"})
String PASSWORD_SPECIAL_CHARS = "##$%^`<>&+=\"!ºª·#~%&'¿¡€,:;*/+-.=_\\[\\]\\(\\)\\|\\_\\?\\\\";
int PASSWORD_MIN_SIZE = 8;
String PASSWORD_REGEXP = "^(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[" + PASSWORD_SPECIAL_CHARS + "])(?=\\S+$).{"+PASSWORD_MIN_SIZE+",}$";
Unit tested:

Password Requirement :
Password should be at least eight (8) characters in length where the system can support it.
Passwords must include characters from at least two (2) of these groupings: alpha, numeric, and special characters.
^.*(?=.{8,})(?=.*\d)(?=.*[a-zA-Z])|(?=.{8,})(?=.*\d)(?=.*[!##$%^&])|(?=.{8,})(?=.*[a-zA-Z])(?=.*[!##$%^&]).*$
I tested it and it works

For anyone interested in minimum requirements for each type of character, I would suggest making the following extension over Tomalak's accepted answer:
^(?=(.*[0-9]){%d,})(?=(.*[a-z]){%d,})(?=(.*[A-Z]){%d,})(?=(.*[^0-9a-zA-Z]){%d,})(?=\S+$).{%d,}$
Notice that this is a formatting string and not the final regex pattern. Just substitute %d with the minimum required occurrences for: digits, lowercase, uppercase, non-digit/character, and entire password (respectively). Maximum occurrences are unlikely (unless you want a max of 0, effectively rejecting any such characters) but those could be easily added as well. Notice the extra grouping around each type so that the min/max constraints allow for non-consecutive matches. This worked wonders for a system where we could centrally configure how many of each type of character we required and then have the website as well as two different mobile platforms fetch that information in order to construct the regex pattern based on the above formatting string.

This one checks for every special character :
^(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=\S+$).*[A-Za-z0-9].{8,}$

Java Method ready for you, with parameters
Just copy and paste and set your desired parameters.
If you don't want a module, just comment it or add an "if" as done by me for special char
//______________________________________________________________________________
/**
* Validation Password */
//______________________________________________________________________________
private static boolean validation_Password(final String PASSWORD_Arg) {
boolean result = false;
try {
if (PASSWORD_Arg!=null) {
//_________________________
//Parameteres
final String MIN_LENGHT="8";
final String MAX_LENGHT="20";
final boolean SPECIAL_CHAR_NEEDED=true;
//_________________________
//Modules
final String ONE_DIGIT = "(?=.*[0-9])"; //(?=.*[0-9]) a digit must occur at least once
final String LOWER_CASE = "(?=.*[a-z])"; //(?=.*[a-z]) a lower case letter must occur at least once
final String UPPER_CASE = "(?=.*[A-Z])"; //(?=.*[A-Z]) an upper case letter must occur at least once
final String NO_SPACE = "(?=\\S+$)"; //(?=\\S+$) no whitespace allowed in the entire string
//final String MIN_CHAR = ".{" + MIN_LENGHT + ",}"; //.{8,} at least 8 characters
final String MIN_MAX_CHAR = ".{" + MIN_LENGHT + "," + MAX_LENGHT + "}"; //.{5,10} represents minimum of 5 characters and maximum of 10 characters
final String SPECIAL_CHAR;
if (SPECIAL_CHAR_NEEDED==true) SPECIAL_CHAR= "(?=.*[##$%^&+=])"; //(?=.*[##$%^&+=]) a special character must occur at least once
else SPECIAL_CHAR="";
//_________________________
//Pattern
//String pattern = "(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[##$%^&+=])(?=\\S+$).{8,}";
final String PATTERN = ONE_DIGIT + LOWER_CASE + UPPER_CASE + SPECIAL_CHAR + NO_SPACE + MIN_MAX_CHAR;
//_________________________
result = PASSWORD_Arg.matches(PATTERN);
//_________________________
}
} catch (Exception ex) {
result=false;
}
return result;
}

Also You Can Do like This.
public boolean isPasswordValid(String password) {
String regExpn =
"^(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[##$%^&+=])(?=\\S+$).{8,}$";
CharSequence inputStr = password;
Pattern pattern = Pattern.compile(regExpn,Pattern.CASE_INSENSITIVE);
Matcher matcher = pattern.matcher(inputStr);
if(matcher.matches())
return true;
else
return false;
}

Use Passay library which is powerful api.

I think this can do it also (as a simpler mode):
^(?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[##$%^&+=])[^\s]{8,}$
[Regex Demo]

easy one
("^ (?=.* [0-9]) (?=.* [a-z]) (?=.* [A-Z]) (?=.* [\\W_])[\\S]{8,10}$")
(?= anything ) ->means positive looks forward in all input string and make sure for this condition is written .sample(?=.*[0-9])-> means ensure one digit number is written in the all string.if not written return false
.
(?! anything ) ->(vise versa) means negative looks forward if condition is written return false.
close meaning ^(condition)(condition)(condition)(condition)[\S]{8,10}$

String s=pwd;
int n=0;
for(int i=0;i<s.length();i++)
{
if((Character.isDigit(s.charAt(i))))
{
n=5;
break;
}
else
{
}
}
for(int i=0;i<s.length();i++)
{
if((Character.isLetter(s.charAt(i))))
{
n+=5;
break;
}
else
{
}
}
if(n==10)
{
out.print("Password format correct <b>Accepted</b><br>");
}
else
{
out.print("Password must be alphanumeric <b>Declined</b><br>");
}
Explanation:
First set the password as a string and create integer set o.
Then check the each and every char by for loop.
If it finds number in the string then the n add 5. Then jump to the
next for loop. Character.isDigit(s.charAt(i))
This loop check any alphabets placed in the string. If its find then
add one more 5 in n. Character.isLetter(s.charAt(i))
Now check the integer n by the way of if condition. If n=10 is true
given string is alphanumeric else its not.

Sample code block for strong password:
(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[^a-zA-Z0-9])(?=\\S+$).{6,18}
at least 6 digits
up to 18 digits
one number
one lowercase
one uppercase
can contain all special characters

RegEx is -
^(?:(?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[##$%^&+=]).*)[^\s]{8,}$
at least 8 digits {8,}
at least one number (?=.*\d)
at least one lowercase (?=.*[a-z])
at least one uppercase (?=.*[A-Z])
at least one special character (?=.*[##$%^&+=])
No space [^\s]

A more general answer which accepts all the special characters including _ would be slightly different:
^(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[\W|\_])(?=\S+$).{8,}$
The difference (?=.*[\W|\_]) translates to "at least one of all the special characters including the underscore".