How do I use regex to select everything before last 4 char in a capture group?
Example:
String str = "{Index1=StudentData(studentName=Sam, idNumber=321231312), Index2=StudentData(studentName=Adam, idNumber=5675), Index3=StudentData(studentName=Lisa, idNumber=67124)}";
String regex = "(?<=idNumber=)[a-zA-Z1-9]+(?=\))";
System.out.println(str.replaceAll(regex, "*"));
Current output:
{Index1=StudentData(studentName=Sam, idNumber=*), Index2=StudentData(studentName=Adam, idNumber=*), Index3=StudentData(studentName=Lisa, idNumber=*)}
Desired output:
{Index1=StudentData(studentName=Sam, idNumber=*****1312), Index2=StudentData(studentName=Adam, idNumber=5675), Index3=StudentData(studentName=Lisa, idNumber=*7124)
You can use this regex in Java:
(\hidNumber=|(?!^)\G)[a-zA-Z1-9](?=[a-zA-Z1-9]{4,}\))
And replace with $1*.
RegEx Demo
Java Code:
final String re = "(\\hidNumber=|(?!^)\\G)[a-zA-Z1-9](?=[a-zA-Z1-9]{4,}\\));
String r = s.replaceAll(re, "$1*");
Breakdown:
(: Start capture group #1
\h: Match a whitespace
idNumber=: Match text idNumber=
|: OR
(?!^)\G: Start at the end of the previous match
): Close capture group #1
[a-zA-Z1-9]: Match an ASCII letter or digit 1-9
(?=[a-zA-Z1-9]{4,}\)): Make sure that ahead of current position we have at least 4 ASCII letters or digits 1-9 followed by )
Related
I am having a string which can have a sentence containing symbols and numbers and also the sentence can have different lengths
For Example
String myString = " () Huawei manufactures phones"
And the next time myString can have the following words
String myString = " * Audi has amazing cars &^"
How can i use regex to get the first word from the string so that the only word i get in the first myString is "Huawei" and the word i get on the second myString is Audi
Below is what i have tried but it fails when there is a space before the first words and symbols
String regexString = myString .replaceAll("\\s.*","")
You may use this regex with a capture group for matching:
^\W*\b(\w+).*
and replace with: $1
RegEx Demo
Java Code:
s = s.replaceAll("^\\W*\\b(\\w+).*", "$1");
RegEx Details:
^: Start
\W*: Match 0 or more non-word characters
\b: Word boundary
(\w+): Match 1+ word characters and capture it in group #1
.*: Match anything aftereards
See how you get on with:
s = s.replaceAll("^[^\\p{Alpha}]*", "");
I want to have a regex for NAME;NAME;NAME and also for NAME;NAME;NAME;NAME where the fourth occurrence of NAME is optional.
I have one regex as (.+);(.+);(.+) which matched the first pattern but not the second. I tried playing with ? but its not working out with (.+);(.+);(.+)(;(.+))?
Basically, I want to achieve the fourth (.+) as zero or one occurence.
Using .+ matches 1+ times any character including ;
If you want to match 3 or 4 groups separated by a ; and not including it, you could use a negated character class [^;]+ with an optional group at the end of the pattern.
^([^;]+);([^;]+);([^;]+)(?:;([^;]+))?$
^ Start of string
([^;]+);([^;]+);([^;]+) Capture group 1, 2 and 3 matching any char except ;
(?: Non capture group
;([^;]+) Match ; and capture any char except ; in group 4
)? Close group and make it optional
$ End of string
Regex demo
If the parts in between can not contain ; you could also use split and count the number of the parts.
String arr[] = { "NAME;NAME;", "NAME;NAME;NAME", "NAME;NAME;NAME;NAME", "NAME;NAME;NAME;NAME;NAME" };
for (String s : arr) {
String [] parts = s.split(";");
if (parts.length == 3 || parts.length == 4) {
System.out.println(s);
}
}
Output
NAME;NAME;NAME
NAME;NAME;NAME;NAME
You can use the regex, (.+);\1;\1(?:;\1)?
Demo:
import java.util.stream.Stream;
public class Main {
public static void main(String args[]) {
// Test
Stream.of(
"NAME;NAME;NAME",
"NAME;NAME;NAME;NAME",
"NAME;NAME;NAME;",
"NAME;NAME;NAMES",
"NAME;NAME;NAME;NAME;NAME"
).forEach(s -> System.out.println(s + " => " + s.matches("(.+);\\1;\\1(?:;\\1)?")));
}
}
Output:
NAME;NAME;NAME => true
NAME;NAME;NAME;NAME => true
NAME;NAME;NAME; => false
NAME;NAME;NAMES => false
NAME;NAME;NAME;NAME;NAME => false
Explanation of the regex:
\1 matches the same text as most recently matched by the 1st capturing group.
?: makes (?:;\1) a non-capturing group.
? makes the previous token optional
With your shown samples, please try following.
1st solution:
^(?:([^;]*);){2,3}\1$
Online demo for 1st solution
Explanation: Adding detailed explanation for above.
^(?: ##Matching value from starting of the value here.
([^;]*); ##Creating 1st capturing group which has everything till ; in it, followed by ;.
){2,3} ##Looking for 2 to 3 occurrences of it.
\1$ ##Again matching 1st capturing group value at the end here.
2nd solution:
^([^;]*)(;)(?:\1\2){1,2}\1$
Online demo for 2nd solution
Explanation: Adding detailed explanation for above.
^([^;]*) ##checking from starting of value, a capturing group till value of ; is coming here.
(;) ##Creating 2nd capturing group which has ; in it.
(?: ##Creating a non-capturing group here.
\1\2 ##Matching 1st and 2nd capturing group here.
){1,2} ##Closing non-capturing group here, with occurrences of 1 to 2.
\1$ ##Matching 1st capturing group value here at the end of value.
You could use lazy quantifier +?. Example:
private static final Pattern pattern = Pattern.compile("((\\w+);?)+?");
public void extractGroups(String input) {
var matcher = pattern.matcher(input);
while (matcher.find()) {
System.out.println(matcher.group(2));
}
}
Input "FIRST;SECOND;THIRD;FOURTH" gives
FIRST
SECOND
THIRD
FOURTH
Input "FIRST;SECOND;THIRD" gives
FIRST
SECOND
THIRD
Lazy quantifier is used to match the shortest possible String. And if you call it repeatedly in while loop, you'll get all matches.
Also you should better use \\w for mathing words, cause . also includes the ; symbol;
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}$)
Start your engine!
I am trying to write a regular expression to mask the below string. Example below.
Input
A1../D//FASDFAS--DFASD//.F
Output (Skip first five and last two Alphanumeric's)
A1../D//FA***********D//.F
I am trying using below regex
([A-Za-z0-9]{5})(.*)(.{2})
Any help would be highly appreciated.
You solve your issue by using Pattern and Matcher with a regex which match multiple groups :
String str = "A1../D//FASDFAS--DFASD//.F";
Pattern pattern = Pattern.compile("(.*?\\/\\/..)(.*?)(.\\/\\/.*)");
Matcher matcher = pattern.matcher(str);
if (matcher.find()) {
str = matcher.group(1)
+ matcher.group(2).replaceAll(".", "*")
+ matcher.group(3);
}
Detail
(.*?\\/\\/..) first group to match every thing until //
(.*?) second group to match every thing between group one and three
(.\\/\\/.*) third group to match every thing after the last character before the // until the end of string
Outputs
A1../D//FA***********D//.F
I think this solution is more readable.
If you want to do that with a single regex you may use
text = text.replaceAll("(\\G(?!^|(?:[0-9A-Za-z][^0-9A-Za-z]*){2}$)|^(?:[^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)
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