Regex Split and Replace string - Java

Question

I am trying to match a regex pattern to replace a particular string.

Sample Text : ABC/1111111031111111/0318*12345678

\/(\d{12,19})\/(0[1-9]|1[0-2])(\d{2})\*

I want to replace the 03 and 18 in /0318 with with "/1222" i.e (Dec-2022). I tried the string replaceAll method but that replaces all the matched characters in the provided sample string. Something like below;

Sample Code i tried:

sampleText.replace(matcher.group(2), "12");

sampleText.replace(matcher.group(3), "22")

How can i do this ?

SAMPLE I/P	EXPECTED O/P
ABC/1111111031111111/0318*12345678	ABC/1111111031111111/1222*12345678
ABC/1852111031156311/1120*12345678	ABC/1852111031156311/1222*12345678

See ideone.com/N6325W, .replaceAll("(/\\d{12,19}/)(?:0[1-9]|1[0-2])\\d{2}(\\*)", "$11222$2") works. — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Apr 29, 2022 at 16:16
Does the string of interest always follow the last forward slash, as in the examples? Aside from doing the substitution do you want the regex to verify the string’s format so that substitutions are only performed for strings having a required format? For example, do you wish to require the string to begin ’ABC’ or that it contain exactly two forward slashes followed by at least so-many digits? I have these questions because you cast your question in terms of examples. Examples are fine for illustration but they are not a substitute for a precise statement of the question in words. — Cary Swoveland
– Cary Swoveland, Commented Apr 29, 2022 at 17:34

The fourth bird · Accepted Answer · 2022-04-29 15:30:57Z

That is because you are replacing the match from group 2 (which is 03) with 12, and replacing the match from group 3 (which is 18) with 22 using String replace() which replaces all occurrences.

Java supports a finite quantifier in a lookbehind assertion using a regex, and you can get a match only to replace with the string 1222 (you don't need 2 separate groups to match the 4 digits).

(?<=/\d{12,19}/)(?:0[1-9]|1[0-2])\d{2}(?=\*)

Explanation

(?<=/\d{12,19}/) Assert 12-19 digits between / to the left
(?:0[1-9]|1[0-2])\d{2} Match 01 - 09 or 1-2 followed by 2 digits
(?=\*) Assert * to the right

See a Java and a regex demo.

Example

String sampleText = "ABC/1111111031111111/0318*12345678\n"
        + "ABC/1852111031156311/1120*12345678";

String pattern = "(?<=/\\d{12,19}/)(?:0[1-9]|1[0-2])\\d{2}(?=\\*)";

System.out.println(sampleText.replaceAll(pattern, "1222"));

Output

ABC/1111111031111111/1222*12345678
ABC/1852111031156311/1222*12345678

Wiktor Stribiżew · Accepted Answer · 2022-04-29 16:16:43Z

1

You can use

(/\d{12,19}/)(?:0[1-9]|1[0-2])\d{2}(\*)

Replace with $11222$2. See the regex demo.

Details:

(/\d{12,19}/) - Group 1: /, 12 to 19 digits
(?:0[1-9]|1[0-2]) - 0 and then a non-zero digit, or a 1 and then 0, 1 or 2
\d{2} - two digits
(\*) - Group 2 ($2): a * char.

See the Java demo

List<String> strs = Arrays.asList(
        "ABC/1111111031111111/0318*12345678", 
        "ABC/1111111031111111/1120*12345678");

String pattern = "(/\\d{12,19}/)(?:0[1-9]|1[0-2])\\d{2}(\\*)";
for (String str : strs)
    System.out.println(str + " => " + str.replaceAll(pattern, "$11222$2"));

Output:

ABC/1111111031111111/0318*12345678 => ABC/1111111031111111/1222*12345678
ABC/1111111031111111/1120*12345678 => ABC/1111111031111111/1222*12345678

edited Apr 29, 2022 at 16:16

answered Apr 29, 2022 at 14:58

Wiktor Stribiżew

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