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Partially mask data of a group of number using regex

I would like to partially mask data using regex. Here is the input :
123-12345-1234567
And here is what I'd like as output :
1**-*****-*****67
I figure out how to replace for the last group but I don't know to do for the rest of the data.
String s = "123-12345-1234567";
System.out.println(s.replaceAll("\\d(?=\\d{2})", "*")); // output is *23-***45-*****67
Also, I'd like to use only regex because I have different type of data, so different type of mask. I don't want to create functions for each type of data.
For example :
AAAAAAAAA // becomes ********AA
12334567 // becomes 123******
Thanks for your help !

We can use the following regex replacement approach:
String input = "123-12345-1234567";
String output = input.substring(0, 1) +
input.substring(1, input.length()-2).replaceAll("\\d", "*") +
input.substring(input.length()-2);
System.out.println(output); // 1**-*****-*****67
Here we concatenate together the first digit, followed by the middle portion with all digits replaced by *, along with the final two digits.
Edit: A pure regex solution, which, however, is more lines of code than the above and might be less performant.
String input = "123-12345-1234567";
String pattern = "^(\\d)(.*)(\\d{2})$";
Pattern r = Pattern.compile(pattern);
Matcher m = r.matcher(input);
if (m.find()) {
String output = m.group(1) + m.group(2).replaceAll("\\d", "*") + m.group(3);
System.out.println(output); // 1**-*****-*****67
}

Java supports a fixed quantifier in a lookbehind, so what you might do is use a pattern with an alternation to account for the different scenario's if you must use a regex only.
Using the lookarounds you can select a single character to be replaced by *
Note that this is hard to maintain, and it would be a better option to write separate functions for the different data formats using separate patterns or string functions (perhaps accompanied by unit tests)
(?<=^\d{3,7})\d(?=\d*$)|(?<=^[A-Z]{0,6})[A-Z](?=[A-Z]*$)|\d(?<=^\d{2,3})(?=\d?-\d{5}-\d{7}$)|\d(?<=^\d{3}-\d{1,5}(?:-\d{1,5})?)
The separate parts match:
(?<=^\d{3,7})\d(?=\d*$) Match a digit asserting 3-7 digits to the left and only digits to the right
| Or
(?<=^[A-Z]{0,6})[A-Z](?=[A-Z]*$) Match A-Z asserting 0-6 chars to the left and only chars A-Z to the right
| Or
\d(?<=^\d{2,3})(?=\d?-\d{5}-\d{7}$) Match a digit asserting 2-3 digits to the left and optional digit, - with 5 digits and - with 7 digits to the right
| Or
\d(?<=^\d{3}-\d{1,5}(?:-\d{1,5})?) Match a digit asserting 3 digits to the left followed - and 1-5 digits and optionally - with 1-5 digits
Regex demo | Java demo
String regex = "(?<=^\\d{3,7})\\d(?=\\d*$)|(?<=^[A-Z]{0,6})[A-Z](?=[A-Z]*$)|\\d(?<=^\\d{2,3})(?=\\d?-\\d{5}-\\d{7}$)|\\d(?<=^\\d{3}-\\d{1,5}(?:-\\d{1,5})?)";
String s1 = "123-12345-1234567";
String s2 = "AAAAAAAAA";
String s3 = "12334567";
System.out.println(s1.replaceAll(regex, "*"));
System.out.println(s2.replaceAll(regex, "*"));
System.out.println(s3.replaceAll(regex, "*"));
Output
1**-*****-*****67
*******AA
123*****

public static void main(String[] args) {
System.out.println("123-12345-1234567".replaceAll("(?<=.{1,})\\d(?=.{3,})", "*"));
System.out.println("AAAAAAAAA".replaceAll(".(?=.{2,})", "*"));
System.out.println("12334567".replaceAll("(?<=.{3,}).", "*"));
}
output:
1**-*****-*****67
*******AA
123*****

Turn zeroes in a string to spaces

Suppose I have a string:
String s = "34205478200044520042";
I want to change this string so that a 0 becomes a space, but consecutive 0s are treated as only 1 space still.
Meaning the new string would look like "342 54782 4452 42"
How do you treat any number of 0s to equal one space?

The regular expression 0+ matches any number of zeroes, because + in a regular expression means "one or more of the preceding subexpression". So you can write
String newString = myString.replaceAll("0+", " ");

You can use the replaceAll method to replace a regular expression of a sequence of 0 characters with a space:
String result = s.replaceAll("0+", " ");

You can use regular expressions to achieve this:
System.out.println(s.replaceAll("[0]+", " "));
// + indicates that the character set (in this case "0") must match at least once.
// Displays 342 54782 4452 42

How do I enter a "." 2 spaces before every "," in a Java string

I've got a string in my Java project which looks something like this
9201,92710,94500,920,1002
How can I enter a dot 2 places before the comma? So it looks like
this:
920.1,9271.0,9450.0,92.0,100.2
I had an attempt at it but I can't get the last number to get a dot.
numbers = numbers.replaceAll("([0-9],)", "\\.$1");
The result I got is
920.1,9271.0,9450.0,92.0,1002
Note: The length of the string is not always the same. It can be longer / shorter.

Check if string ends with ",". If not, append a "," to the string, run the same replaceAll, remove "," from end of String.
Split string by the "," delimiter, process each piece adding the "." where needed.
Just add a "." at numbers.length-1 to solve the issue with the last number

As your problem is not only inserting the dot before every comma, but also before end of string, you just must add this additional condition to your capturing group:
numbers = numbers.replaceAll("([0-9](,|$))", "\\.$1");
As suggested by Siguza, you could as well use a non-capturing group which is even more what a "human" would expect to be captured in the capturing group:
numbers = numbers.replaceAll("([0-9](?:,|$))", "\\.$1");
But as a non-capturing group is (although a really nice feature) not standard Regex and the overhead is not that significant here, I would recommend using the first option.

You could use word boundary:
numbers = numbers.replaceAll("(\\d)\b", ".$1");

Your solution is fine, as long as you put a comma at the end like dan said.
So instead of:
numbers = numbers.replaceAll("([0-9],)", "\\.$1");
write:
numbers = (numbers+",").replaceAll("([0-9],)", "\\.$1");
numbers = numbers.substring(0,numbers.size()-1);

You may use a positive lookahead to check for the , or end of string right after a digit and a zeroth backreference to the whole match:
String s = "9201,92710,94500,920,1002";
System.out.println(s.replaceAll("\\d(?=,|$)", ".$0"));
// => 920.1,9271.0,9450.0,92.0,100.2
See the Java demo and a regex demo.
Details:
\\d - exactly 1 digit...
(?=,|$) - that must be before a , or end of string ($).
A capturing variation (Java demo):
String s = "9201,92710,94500,920,1002";
System.out.println(s.replaceAll("(\\d)(,|$)", ".$1$2"));

You where right to go for the replaceAll method. But your regex was not matching the end of the string, the last set of numbers.
Here is my take on your problem:
public static void main(String[] args) {
String numbers = "9201,92710,94500,920,1002";
System.out.println(numbers.replaceAll("(\\d,|\\d$)", ".$1"));
}
the regex (\\d,|\\d$) matches a digit followed by a comma \d,, OR | a digit followed by the end of the string \d$.
I have tested it and found to work.
As others have suggested you could add a comma at the end, run the replace all and then remove it. But it seems as extra effort.
Example:
public static void main(String[] args) {
String numbers = "9201,92710,94500,920,1002";
//add on the comma
numbers += ",";
numbers = numbers.replaceAll("(\\d,)", "\\.$1");
//remove the comma
numbers = numbers.substring(0, numbers.length()-1);
System.out.println(numbers);
}

How can I strip all non digits in a string except the first character?

I have a string that I want to make sure that the format is always a + followed by digits.
The following would work:
String parsed = inputString.replaceAll("[^0-9]+", "");
if(inputString.charAt(0) == '+') {
result = "+" + parsed;
}
else {
result = parsed;
}
But is there a way to have a regex in the replaceAll that would keep the + (if exists) in the beginning of the string and replace all non digits in the first line?

The following statement with the given regex would do the job:
String result = inputString.replaceAll("(^\\+)|[^0-9]", "$1");
(^\\+) find either a plus sign at the beginning of string and put it to a group ($1),
| or
[^0-9] find a character which is not a number
$1 and replace it with nothing or the plus sign at the start of group ($1)

You can use this expression:
String r = s.replaceAll("((?<!^)[^0-9]|^[^0-9+])", "");
The idea is to replace any non-digit when it is not the initial character of the string (that's the (?<!^)[^0-9] part with a lookbehind) or any character that is not a digit or plus that is the initial character of the string (the ^[^0-9+] part).
Demo.

What about just
(?!^)\D+
Java string:
"(?!^)\\D+"
Demo at regex101.com
\D matches a character that is not a digit [^0-9]
(?!^) using a negative lookahead to check, if it is not the initial character

Yes you can use this kind of replacement:
String parsed = inputString.replaceAll("^[^0-9+]*(\\+)|[^0-9]+", "$1");
if present and before the first digit in the string, the + character is captured in group 1. For example: dfd+sdfd12+sdf12 returns +1212 (the second + is removed since its position is after the first digit).

try this
1- This will allow negative and positive number and will match app special char except - and + at first position.
(?!^[-+])[^0-9.]
2- If you only want to allow + at first position
(?!^[+])[^0-9.]

Regular Expression - inserting space after comma only if succeeded by a letter or number

In Java I want to insert a space after a String but only if the character after the comma is succeeded by a digit or letter. I am hoping to use the replaceAll method which uses regular expressions as a parameter. So far I have the following:
String s1="428.0,chf";
s1 = s1.replaceAll(",(\\d|\\w)",", ");
This code does successfully distinguish between the String above and one where there is already a space after the comma. My problem is that I can't figure out how to write the expression so that the space is inserted. The code above will replace the c in the String shown above with a space. This is not what I want.
s1 should look like this after executing the replaceAll: "428.0 chf"

s1.replaceAll(",(?=[\da-zA-Z])"," ");
(?=[\da-zA-Z]) is a positive lookahead which would look for a digit or a word after ,.This lookahead would not be replaced since it is never included in the result.It's just a check
NOTE
\w includes digit,alphabets and a _.So no need of \d.
A better way to represent it would be [\da-zA-Z] instead of \w since \w also includes _ which you do not need 2 match

Try this, and note that $1 refers to your matched grouping:
s1.replaceAll(",(\\d|\\w)"," $1");
Note that String.replaceAll() works in the same way as a Matcher.replaceAll(). From the doc:
The replacement string may contain references to captured subsequences

String s1="428.0,chf";
s1 = s1.replaceAll(",([^_]\\w)"," $1"); //Match alphanumeric except '_' after ','
System.out.println(s1);
Output: -
428.0 chf
Since \w matches digits, words, and an underscore, So, [^_] negates the underscore from \w..
$1 represents the captured group.. You captured c after , here, so replace c with _$1 -> _c.. "_" represent a space..

Try this....
public class Tes {
public static void main(String[] args){
String s1="428.0,chf";
String[] sArr = s1.split(",");
String finalStr = new String();
for(String s : sArr){
finalStr = finalStr +" "+ s;
}
System.out.println(finalStr);
}
}