I am trying to write a regular expression to mask the below string. Example below.
Input
A1../D//FASDFAS--DFASD//.F
Output (Skip first five and last two Alphanumeric's)
A1../D//FA***********D//.F
I am trying using below regex
([A-Za-z0-9]{5})(.*)(.{2})
Any help would be highly appreciated.
You solve your issue by using Pattern and Matcher with a regex which match multiple groups :
String str = "A1../D//FASDFAS--DFASD//.F";
Pattern pattern = Pattern.compile("(.*?\\/\\/..)(.*?)(.\\/\\/.*)");
Matcher matcher = pattern.matcher(str);
if (matcher.find()) {
str = matcher.group(1)
+ matcher.group(2).replaceAll(".", "*")
+ matcher.group(3);
}
Detail
(.*?\\/\\/..) first group to match every thing until //
(.*?) second group to match every thing between group one and three
(.\\/\\/.*) third group to match every thing after the last character before the // until the end of string
Outputs
A1../D//FA***********D//.F
I think this solution is more readable.
If you want to do that with a single regex you may use
text = text.replaceAll("(\\G(?!^|(?:[0-9A-Za-z][^0-9A-Za-z]*){2}$)|^(?:[^0-9A-Za-z]*[0-9A-Za-z]){5}).", "$1*");
Or, using the POSIX character class Alnum:
text = text.replaceAll("(\\G(?!^|(?:\\p{Alnum}\\P{Alnum}*){2}$)|^(?:\\P{Alnum}*\\p{Alnum}){5}).", "$1*");
See the Java demo and the regex demo. If you plan to replace any code point rather than a single code unit with an asterisk, replace . with \P{M}\p{M}*+ ("\\P{M}\\p{M}*+").
To make . match line break chars, add (?s) at the start of the pattern.
Details
(\G(?!^|(?:[0-9A-Za-z][^0-9A-Za-z]*){2}$)|^(?:[^0-9A-Za-z]*[0-9A-Za-z]){5}) -
\G(?!^|(?:[0-9A-Za-z][^0-9A-Za-z]*){2}$) - a location after the successful match that is not followed with 2 occurrences of an alphanumeric char followed with 0 or more chars other than alphanumeric chars
| - or
^(?:[^0-9A-Za-z]*[0-9A-Za-z]){5} - start of string, followed with five occurrences of 0 or more non-alphanumeric chars followed with an alphanumeric char
. - any code unit other than line break characters (if you use \P{M}\p{M}*+ - any code point).
Usually, masking of characters in the middle of a string can be done using negative lookbehind (?<!) and positive lookahead groups (?=).
But in this case lookbehind group can't be used because it does not have an obvious maximum length due to unpredictable number of non-alphanumeric characters between first five alphanumeric characters (. and / in the A1../D//FA).
A substring method can used as a workaround for inability to use negative lookbehind group:
String str = "A1../D//FASDFAS--DFASD//.F";
int start = str.replaceAll("^((?:\\W{0,}\\w{1}){5}).*", "$1").length();
String maskedStr = str.substring(0, start) +
str.substring(start).replaceAll(".(?=(?:\\W{0,}\\w{1}){2})", "*");
System.out.println(maskedStr);
// A1../D//FA***********D//.F
But the most straightforward way is to use java.util.regex.Pattern and java.util.regex.Matcher:
String str = "A1../D//FASDFAS--DFASD//.F";
Pattern pattern = Pattern.compile("^((?:\\W{0,}\\w{1}){5})(.+)((?:\\W{0,}\\w{1}){2})");
Matcher matcher = pattern.matcher(str);
if (matcher.find()) {
String maskedStr = matcher.group(1) +
"*".repeat(matcher.group(2).length()) +
matcher.group(3);
System.out.println(maskedStr);
// A1../D//FA***********D//.F
}
\W{0,} - 0 or more non-alphanumeric characters
\w{1} - exactly 1 alphanumeric character
(\W{0,}\w{1}){5} - 5 alphanumeric characters and any number of alphanumeric characters in between
(?:\W{0,}\w{1}){5} - do not capture as a group
^((?:\\W{0,}\\w{1}){5})(.+)((?:\\W{0,}\\w{1}){2})$ - substring with first five alphanumeric characters (group 1), everything else (group 2), substring with last 2 alphanumeric characters (group 3)
Related
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!
I have the following method I'm trying to implement: parses the input into “word tokens”: sequences of word characters separated by non-word characters. However, non-word characters can become part of a token if they are quoted (in single quotes).
I want to use regex but have trouble getting my code just right:
public static List<String> wordTokenize(String input) {
Pattern pattern = Pattern.compile ("\\b(?:(?<=\')[^\']*(?=\')|\\w+)\\b");
Matcher matcher = pattern.matcher (input);
ArrayList ans = new ArrayList();
while (matcher.find ()){
ans.add (matcher.group ());
}
return ans;
}
My regex fails to identify that starting a word mid word without space doesn't mean starting a new word. Examples:
The input: this-string 'has only three tokens' // works
The input:
"this*string'has only two#tokens'"
Expected :[this, stringhas only two#tokens]
Actual :[this, string, has only two#tokens]
The input: "one'two''three' '' four 'twenty-one'"
Expected :[onetwothree, , four, twenty-one]
Actual :[one, two, three, four, twenty-one]
How do I fix the spaces?
You want to match one or more occurrences of a word char or a substring between the closest single straight apostrophes, and remove all those apostrophes from the tokens.
Use the following regex and .replace("'", "") on the matches:
(?:\w|'[^']*')+
See the regex demo. Details:
(?: - start of a non-capturing group
\w - a word char
| - or
' - a straight single quotation mark
[^']* - any 0+ chars other than a straight single quotation mark
' - a straight single quotation mark
)+ - end of the group, 1+ occurrences.
See the Java demo:
// String s = "this*string'has only two#tokens'"; // => [this, stringhas only two#tokens]
String s = "one'two''three' '' four 'twenty-one'"; // => [onetwothree, , four, twenty-one]
Pattern pattern = Pattern.compile("(?:\\w|'[^']*')+", Pattern.UNICODE_CHARACTER_CLASS);
Matcher matcher = pattern.matcher(s);
List<String> tokens = new ArrayList<>();
while (matcher.find()){
tokens.add(matcher.group(0).replace("'", ""));
}
Note the Pattern.UNICODE_CHARACTER_CLASS is added for the \w pattern to match all Unicode letters and digits.
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 ($).
This works like you have described
public class MyRegex{
public static void main(String[] args) {
Pattern patern = Pattern.compile("#(\\d++)[^\\.]");
Matcher matcher1 = patern.matcher("my car number is #8746253 which is actually cool");
if(matcher1.find()){
System.out.println(matcher1.group(1));
}
Matcher matcher2 = patern.matcher("earth is #8746253.Kms away, which is very far");
if(matcher2.find()){
System.out.println(matcher1.group(1));
}else{
System.out.println("No match found");
}
}
}
Outputs:
> 8746253
> No match found
I am very new to regex, I am learning it now. I have a requirement like this:
Any String starts with #newline# and also ends with #newline#. In between these two words, there could be (0 or more spaces) or (0 or more #newline#).
below is an example:
#newline# #newline# #newline##newline# #newline##newline##newline#.
How to do regex for this?
I have tried this, but not working
^#newline#|(\s+#newline#)|#newline#|#newline#$
Your ^#newline#|(\s+#newline#)|#newline#|#newline#$ matches either a #newline# at the start of the string (^#newline#), or 1+ whitespaces followed with #newline# ((\s+#newline#)), or #newline#, or (and this never matches as the previous catches all the cases of #newline#) a #newline# at the end of the string (#newline#$).
You may match these strings with
^#newline#(?:\s*#newline#)*$
or (if there should be at least 2 occurrences of #newline# in the string)
^#newline#(?:\s*#newline#)+$
^
See the regex demo.
^ - start of string
#newline# - literal string
(?:\s*#newline#)* - zero (NOTE: replacing * with + will require at least 1) or more sequences of
\s* - 0+ whitespaces
#newline# - a literal substring
$ - end of string.
Java demo:
String s = "#newline# #newline# #newline##newline# #newline##newline##newline#";
System.out.println(s.matches("#newline#(?:\\s*#newline#)+"));
// => true
Note: inside matches(), the expression is already anchored, and ^ and $ can be removed.
As far as I understand the requirements, it should be this:
^#newline#(\s|#newline#)*#newline#$
this will not match your example string, since it does not start with #newline#
without the ^ and the $ it matches a sub-string.
Check out http://www.regexplanet.com/ to play around with Regular Expressions.
Please use the pattern and matches classes to identify.
You can give the patternString string at runtime
patternString="newline";
public void findtheMatch(String patternString)
{
String text ="#newline# #newline# #newline##newline# #newline##newline##newline# ";
Pattern pattern = Pattern.compile(patternString);
Matcher matcher = pattern.matcher(text);
while(matcher.find()) {
System.out.println("found: " + matcher.group(1));
}
}
You can try this as well:
#newline#[\s\S]+#newline#
It says, match anything that starts with #newline# followed by any combination of whitespace or non-whitespace characters and ends with #newline#.
I am wondering what the regex for a word would be, I can seem to find it anywhere? The string I\m trying to match "Loop-num + 5" and I want to extract the "Loop-num" part. I am unsure what the regex would be to do so.
Pattern pattern = Pattern.compile("(loop-.*)");
Matcher matcher = pattern.matcher("5 * loop-num + 5");
if(matcher.find()){
String extractedString = matcher.group(1);
System.out.println(extractedString);
}
From this I get: "loop-num + 5"
If you really plan to use the regex to match words (entities comprising just letters, optionally split with hyphen(s)), you need to consider the following regex:
\b\pL+(?:-\pL+)*\b
See regex demo
Explanation:
\b - leading word boundary
\pL+ - 1 or more Unicode letters
(?:-\pL+)* - zero or more sequences of...
- - a literal hyphen
\pL+ - 1 or more Unicode letters
\b - trailing word boundary
In Java:
Pattern pattern = Pattern.compile("\\b\\pL+(?:-\\pL+)*\\b", Pattern.UNICODE_CHARACTER_CLASS);
Matcher matcher = pattern.matcher("5 * loop-num + 5");
if(matcher.find()){
String extractedString = matcher.group(0);
System.out.println(extractedString);
}
Note: in case words may include digits (not at the starting positions), you can use \b\pL\w*(?:-\pL\w*)*\b with Pattern.UNICODE_CHARACTER_CLASS. Here, \w will match letters, digits and an underscore.