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Masking credit card number using regex

I am trying to mask the CC number, in a way that third character and last three characters are unmasked.
For eg.. 7108898787654351 to **0**********351
I have tried (?<=.{3}).(?=.*...). It unmasked last three characters. But it unmasks first three also.
Can you throw some pointers on how to unmask 3rd character alone?

You can use this regex with a lookahead and lookbehind:
str = str.replaceAll("(?<!^..).(?=.{3})", "*");
//=> **0**********351
RegEx Demo
RegEx Details:
(?<!^..): Negative lookahead to assert that we don't have 2 characters after start behind us (to exclude 3rd character from matching)
.: Match a character
(?=.{3}): Positive lookahead to assert that we have at least 3 characters ahead

I would suggest that regex isn't the only way to do this.
char[] m = new char[16]; // Or whatever length.
Arrays.fill(m, '*');
m[2] = cc.charAt(2);
m[13] = cc.charAt(13);
m[14] = cc.charAt(14);
m[15] = cc.charAt(15);
String masked = new String(m);
It might be more verbose, but it's a heck of a lot more readable (and debuggable) than a regex.

Here is another regular expression:
(?!(?:\D*\d){14}$|(?:\D*\d){1,3}$)\d
See the online demo
It may seem a bit unwieldy but since a credit card should have 16 digits I opted to use negative lookaheads to look for an x amount of non-digits followed by a digit.
(?! - Negative lookahead
(?: - Open 1st non capture group.
\D*\d - Match zero or more non-digits and a single digit.
){14} - Close 1st non capture group and match it 14 times.
$ - End string ancor.
| - Alternation/OR.
(?: - Open 2nd non capture group.
\D*\d - Match zero or more non-digits and a single digit.
){1,3} - Close 2nd non capture group and match it 1 to 3 times.
$ - End string ancor.
) - Close negative lookahead.
\d - Match a single digit.
This would now mask any digit other than the third and last three regardless of their position (due to delimiters) in the formatted CC-number.

Apart from where the dashes are after the first 3 digits, leave the 3rd digit unmatched and make sure that where are always 3 digits at the end of the string:
(?<!^\d{2})\d(?=[\d-]*\d-?\d-?\d$)
Explanation
(?<! Negative lookbehind, assert what is on the left is not
^\d{2} Match 2 digits from the start of the string
) Close lookbehind
\d Match a digit
(?= Positive lookahead, assert what is on the right is
[\d-]* 0+ occurrences of either - or a digit
\d-?\d-?\d Match 3 digits with optional hyphens
$ End of string
) Close lookahead
Regex demo | Java demo
Example code
String regex = "(?<!^\\d{2})\\d(?=[\\d-]*\\d-?\\d-?\\d$)";
Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
String strings[] = { "7108898787654351", "7108-8987-8765-4351"};
for (String s : strings) {
Matcher matcher = pattern.matcher(s);
System.out.println(matcher.replaceAll("*"));
}
Output
**0**********351
**0*-****-****-*351

Don't think you should use a regex to do what you want. You could use StringBuilder to create the required string
String str = "7108-8987-8765-4351";
StringBuilder sb = new StringBuilder("*".repeat(str.length()));
for (int i = 0; i < str.length(); i++) {
if (i == 2 || i >= str.length() - 3) {
sb.replace(i, i + 1, String.valueOf(str.charAt(i)));
}
}
System.out.print(sb.toString()); // output: **0*************351

You may add a ^.{0,1} alternative to allow matching . when it is the first or second char in the string:
String s = "7108898787654351"; // **0**********351
System.out.println(s.replaceAll("(?<=.{3}|^.{0,1}).(?=.*...)", "*"));
// => **0**********351
The regex can be written as a PCRE compliant pattern, too: (?<=.{3}|^|^.).(?=.*...).
The regex can be written as a PCRE compliant pattern, too: (?<=.{3}|^|^.).(?=.*...).
It is equal to
System.out.println(s.replaceAll("(?<!^..).(?=.*...)", "*"));
See the Java demo and a regex demo.
Regex details
(?<=.{3}|^.{0,1}) - there must be any three chars other than line break chars immediately to the left of the current location, or start of string, or a single char at the start of the string
(?<!^..) - a negative lookbehind that fails the match if there are any two chars other than line break chars immediately to the left of the current location
. - any char but a line break char
(?=.*...) - there must be any three chars other than line break chars immediately to the right of the current location.

If the CC number always has 16 digits, as it does in the example, and as do Visa and MasterCard CC's, matches of the following regular expression can be replaced with an asterisk.
\d(?!\d{0,2}$|\d{13}$)
Start your engine!

How to give range in lookhead using regex e.g ^(?=(.*[a-z]){1,3})(?=.*[0-9]).{2,5}$

Question-
1]String length is 2 to 5
2]String contains at least 1 char and maximum 3 char
3]Atleast one number
I want do using lookhead.
What i tried but not working
^(?=(.[a-z]){1,3})(?=.[0-9]).{2,5}$

Try using below regex
(?=([\\w\\W]{2,5}$))(?=(.*([0-9]){1,}.*))^[^A-Za-z]*[A-Za-z](?!(?:[^A-Za-z]*[A-Za-z]){3})
public static void main(String[] args) {
String regex = "(?=([\\w\\W]{2,5}$))(?=(.*([0-9]){1,}.*))^[^A-Za-z]*[A-Za-z](?!(?:[^A-Za-z]*[A-Za-z]){3}).*";
Pattern pattern = Pattern.compile(regex);
System.out.println(pattern.matcher("AB2").find()); // output true
System.out.println(pattern.matcher("AB2C").find()); // output true
}

You might use a positive lookahead to assert that the length of the string is 2-5 characters and that there is at least 1 digit and 1 character [a-z]. To make sure that there is a maximum of 3 characters to can use a negative lookahead to assert that you don't match a character 4 times.
^(?=[a-z0-9]{2,5}$)(?=.*[a-z])(?=.*[0-9])(?!(?:.*[a-z]){4})[a-z0-9]+$
Regex demo
Explanation
^ Assert the start of the line
(?=[a-z0-9]{2,5}$ Positive lookahead to assert 2 -5 characters
(?=.*[a-z]) Positive lookahead to assert 1 character
(?=.*[0-9]) Positive lookahead to assert 1 digit
(?!(?:.*[a-z]){4}) Negative lookahead to not match a character 4 times
[a-z0-9]+$ Match one or more times a character or a digit until the end of the string
Demo

Regular expression for phrase contain literals and numbers but is not all phrase as a number only with fixed range length

i want to have regular expression to check input character as a-z and 0-9 but i do not want to allow input as just numeric value at all ( must be have at least one alphabetic character)
for example :
413123123123131
not allowed but if have just only one alphabetic character in any place of phrase it's ok
i trying to define correct Regex for that and at final i raised to
[0-9]*[a-z].*
but in now i confused how to defined {x,y} length of phrase i want to have {9,31} but after last * i can not to have length block too i trying to define group but unlucky and not worked
tested at https://www.debuggex.com/
how can i to add it ??

What you seek is
String regex = "(?=.{9,31}$)\\p{Alnum}*\\p{Alpha}\\p{Alnum}*";
Use it with String#matches() / Pattern#matches() method to require a full string match:
if (s.matches(regex)) {
return true;
}
Details
^ - implicit in matches() - matches the start of string
(?=.{9,31}$) - a positive lookahead that requires 9 to 31 any chars other than line break chars from the start to end of the string
\\p{Alnum}* - 0 or more alphanumeric chars
\\p{Alpha} - an ASCII letter
\\p{Alnum}* - 0 or more alphanumeric chars
Java demo:
String lines[] = {"413123123123131", "4131231231231a"};
Pattern p = Pattern.compile("(?=.{9,31}$)\\p{Alnum}*\\p{Alpha}\\p{Alnum}*");
for(String line : lines)
{
Matcher m = p.matcher(line);
if(m.matches()) {
System.out.println(line + ": MATCH");
} else {
System.out.println(line + ": NO MATCH");
}
}
Output:
413123123123131: NO MATCH
4131231231231a: MATCH

This might be what you are looking for.
[0-9a-zA-Z]*[a-zA-Z][0-9a-zA-Z]*
To help explain it, think of the middle term as your one required character and the outer terms as any number of alpha numeric characters.
Edit: to restrict the length of the string as a whole you may have to check that manually after matching. ie.
if (str.length > 9 && str.length < 31)
Wiktor does provide a solution that involves more regex, please look at his for a better regex pattern

Try this Regex:
^(?:(?=[a-z])[a-z0-9]{9,31}|(?=\d.*[a-z])[a-z0-9]{9,31})$
OR a bit shorter form:
^(?:(?=[a-z])|(?=\d.*[a-z]))[a-z0-9]{9,31}$
Demo
Explanation(for the 1st regex):
^ - position before the start of the string
(?=[a-z])[a-z0-9]{9,31} means If the string starts with a letter, then match Letters and digits. minimum 9 and maximum 31
| - OR
(?=\d.*[a-z])[a-z0-9]{9,31} means If the string starts with a digit followed by a letter somewhere in the string, then match letters and digits. Minimum 9 and Maximum 31. This also ensures that If the string starts with a digit and if there is no letter anywhere in the string, there won't be any match
$ - position after the last literal of the string
OUTPUT:
413123123123131 NO MATCH(no alphabets)
kjkhsjkf989089054835werewrew65 MATCH
kdfgfd4374985794379857984379857weorjijuiower NO MATCH(length more than 31)
9087erkjfg9080980984590p465467 MATCH
4131231231231a MATCH
kjdfg34 NO MATCH(Length less than 9)

Here's the regex:
[a-zA-Z\d]*[a-zA-Z][a-zA-Z\d]*
The trick here is to have something that is not optional. The leading and trailing [a-zA-Z\d] has a * quantifier, so they are optional. But the [a-zA-Z] in the middle there is not optional. The string must have a character that matches [a-zA-Z] in order to be matched.
However, you need to check the length of the string with length afterwards and not with regex. I can't think of any way how you can do this in regex.
Actually, I think you can do this regexless pretty easily:
private static boolean matches(String input) {
for (int i = 0 ; i < input.length() ; i++) {
if (Character.isLetter(input.charAt(i))) {
return input.length() >= 9 && input.length() <= 31;
}
}
return false;
}

Regex to match a digit not followed by a dot(".")

I have a string
string 1(excluding the quotes) -> "my car number is #8746253 which is actually cool"
conditions - The number 8746253, could be of any length and
- the number can also be immediately followed by an end-of-line.
I want to group-out 8746253 which should not be followed by a dot "."
I have tried,
.*#(\d+)[^.].*
This will get me the number for sure, but this will match even if there is a dot, because [.^] will match the last digit of the number(for example, 3 in the below case)
string 2(excluding the quotes) -> "earth is #8746253.Kms away, which is very far"
I want to match only the string 1 type and not the string 2 types.

To match any number of digits after # that are not followed with a dot, use
(?<=#)\d++(?!\.)
The ++ is a possessive quantifier that will make the regex engine only check the lookahead (?!\.) only after the last matched digit, and won't backtrack if there is a dot after that. So, the whole match will get failed if there is a dit after the last digit in a digit chunk.
See the regex demo
To match the whole line and put the digits into capture group #1:
.*#(\d++)(?!\.).*
See this regex demo. Or a version without a lookahead:
^.*#(\d++)(?:[^.\r\n].*)?$
See another demo. In this last version, the digit chunk can only be followed with an optional sequence of a char that is not a ., CR and LF followed with any 0+ chars other than line break chars ((?:[^.\r\n].*)?) and then the end of string ($).

This works like you have described
public class MyRegex{
public static void main(String[] args) {
Pattern patern = Pattern.compile("#(\\d++)[^\\.]");
Matcher matcher1 = patern.matcher("my car number is #8746253 which is actually cool");
if(matcher1.find()){
System.out.println(matcher1.group(1));
}
Matcher matcher2 = patern.matcher("earth is #8746253.Kms away, which is very far");
if(matcher2.find()){
System.out.println(matcher1.group(1));
}else{
System.out.println("No match found");
}
}
}
Outputs:
> 8746253
> No match found

Regular Expression (RegEx) for User Name in Java

How to form the RegEx of user name string in Java?
Rules in Exercise :
Only 3 - 10 characters.
Only 'a'-'z', 'A'-'Z', '1'-'9', '_' and '.' are allowed.
'_' and '.' can only be appeared 0 to 2 times.
"abc_._" = false
"abc..." = false
"abc__" = true
"abc.." = true
"abc_." = true
If I do not use Regex, it will be easier.
Without considering '1'-'9', I have tried the following RegEx but they are not workable.
String username_regex = "[a-zA-Z||[_||.]{0,2}]{3,10}";
String username_regex = "[a-zA-Z]{3,10}||[_||.]{0,2}";
My function :
public static boolean isUserNameCorrect(String user_name) {
String username_regex = "[a-zA-Z||[_]{0,2}]{3,10}";
boolean isMatch = user_name.matches(username_regex);
return isMatch;
}
What RegEx should I use?

If I remember well from CS classes, it is not possible to create one single regex to satisfy all three requirements. So, I would make separate checks for each condintion. For example, this regex checks for conditions 1 and 2, and condition 3 is checked separately.
private static final Pattern usernameRegex = Pattern.compile("[a-zA-Z1-9._]{3,10}");
public static boolean isUserNameCorrect(String userName) {
boolean isMatch = usernameRegex.matcher(userName).matches();
return isMatch && countChar(userName, '.')<=2 && countChar(userName, '_') <=2;
}
public static int countChar(String s, char c) {
int count = 0;
int index = s.indexOf(c, 0);
while ( index >= 0 ) {
count++;
index = s.indexOf(c, index+1);
}
return count;
}
BTW, notice the pattern that allows you to reuse a regex in Java (performace gain, because it is expensive to compile a regex).
The reason that a regex cannot do what you want (again if I remember well) is that this problem requires a context-free-grammar, while regex is a regular grammar. Ream more

First off, || isn't necessary for this problem, and in fact doesn't do what you think it does. I've only ever seen it used in groups for regex (like if you want to match Hello or World, you'd match (Hello|World) or (?:Hello|World), and in those cases you only use a single |.
Next, let me explain why each of the regex you have tried won't work.
String username_regex = "[a-zA-Z||[_||.]{0,2}]{3,10}";
Range operators inside a character class aren't interpreted as range operators, and instead will just represent the literals that make up the range operators. In addition, nested character classes are simply combined. So this is effectively equal to:
String username_regex = "[a-zA-Z_|.{0,2}]{3,10}";
So it'll match some combination of 3-10 of the following: a-z, A-Z, 0, 2, {, }, ., |, and _.
And that's not what you wanted.
String username_regex = "[a-zA-Z]{3,10}||[_||.]{0,2}";
This will match 3 to 10 of a-z or A-Z, followed by two pipes, followed by _, |, or . 0 to 2 times. Also not what you wanted.
The easy way to do this is by splitting the requirements into two sections and creating two regex strings based off of those:
Only 3 - 10 characters, where only 'a'-'z', 'A'-'Z', '1'-'9', '_' and '.' are allowed.
'_' and '.' can only appear 0 to 2 times.
The first requirement is quite simple: we just need to create a character class including all valid characters and place limits on how many of those can appear:
"[a-zA-Z1-9_.]{3,10}"
Then I would validate that '_' and '.' appear 0 to 2 times:
".*[._].*[._].*"
or
"(?:.*[._].*){0,2}" // Might work, might not. Preferable to above regex if easy configuration is necessary. Might need reluctant quantifiers...
I'm unfortunately not experienced enough to figure out what a single regex would look like... But these are at least quite readable.

May not be elegant but you may try this:
^(([A-Za-z0-9\._])(?!.*[\._].*[\._].*[\._])){3,10}$
Here is the explanation:
NODE EXPLANATION
--------------------------------------------------------------------------------
^ the beginning of the string
--------------------------------------------------------------------------------
( group and capture to \1 (between 3 and 10
times (matching the most amount
possible)):
--------------------------------------------------------------------------------
( group and capture to \2:
--------------------------------------------------------------------------------
[A-Za-z0-9\._] any character of: 'A' to 'Z', 'a' to
'z', '0' to '9', '\.', '_'
--------------------------------------------------------------------------------
) end of \2
--------------------------------------------------------------------------------
(?! look ahead to see if there is not:
--------------------------------------------------------------------------------
.* any character except \n (0 or more
times (matching the most amount
possible))
--------------------------------------------------------------------------------
[\._] any character of: '\.', '_'
--------------------------------------------------------------------------------
.* any character except \n (0 or more
times (matching the most amount
possible))
--------------------------------------------------------------------------------
[\._] any character of: '\.', '_'
--------------------------------------------------------------------------------
.* any character except \n (0 or more
times (matching the most amount
possible))
--------------------------------------------------------------------------------
[\._] any character of: '\.', '_'
--------------------------------------------------------------------------------
) end of look-ahead
--------------------------------------------------------------------------------
){3,10} end of \1 (NOTE: because you are using a
quantifier on this capture, only the LAST
repetition of the captured pattern will be
stored in \1)
--------------------------------------------------------------------------------
$ before an optional \n, and the end of the
string
This will satisfy your above-mentioned requirement. Hope it helps :)

Please try this:
[[a-Z][0-9][._]?[[a-Z][0-9][._]?[[a-Z][0-9]*
Niko
EDIT :
You're right. Then several Regexp :
Regex1: ^[\w.]{3-10}$
Regex2: ^[[a-Z][0-9]][_.]?[[a-Z][0-9]][_.]?[[a-Z][0-9]]*$
I hope I forgot nothing!