使用正则表达式从给定的字符串值中提取子字符串

Question

I have a requirement where a string needs to be matched and then extract further value from a that string

I will receive a header in request whose value will be a DN name from ssl certificate. Here need to match a specific string 1.2.3.47 in the header and extract remaining text.

Sample String passed to method: O=ABC Bank Plc/1.2.3.47=ABC12-PQR-121878, CN=7ltM2wQ3bqlDJdBEURGAMq, L=INDIA, C=INDIA, E=xyz@gmail.com

Here is my code:

private String extractDN(String dnHeader) {
     if(!ValidatorUtil.isEmpty(dnHeader)){

       String tokens[]=dnHeader.split(",");
       if(tokens[0].contains("1.2.3.47")){

          int index=tokens[0].lastIndexOf("1.2.3.47");
          String id=tokens[0].substring(index+9);
          System.out.println(id);


       }



     }
    return id;
  }

Can a regex pattern be used here to match and extract value? Is there any better way to achieve this? Please help.

Answer 1

If you want to use a pattern and if you know that the value always starts with a forward slash and if followed by one or more digits separated by a dot and then an equals sign, you could use a capturing group:

/[0-9](?:\\.[0-9]+)+=([^,]+)

• / Match /
• [0-9]+ Match 1+ digit 0-9
• (?: Non capturing group
  • \\\\.[0-9]+ match . and 1+ digits 0-9
• )+ Close non capturing group and repeat 1+ times
• = Match =
• ([^,]+) Capture group 1, match 1+ times any char except a ,

Regex demo | Java demo

For example

final String regex = "/[0-9]+(?:\\.[0-9]+)+=([^,]+)";
final String string = "O=ABC Bank Plc/1.2.3.47=ABC12-PQR-121878, CN=7ltM2wQ3bqlDJdBEURGAMq, L=INDIA, C=INDIA, E=xyz@gmail.com";

final Pattern pattern = Pattern.compile(regex);
final Matcher matcher = pattern.matcher(string);

if (matcher.find()) {
    System.out.println(matcher.group(1));
}

Output

ABC12-PQR-121878

---

If you want a more precise match, you could also specify the start of the pattern:

\\bO=\\w+(?:\\h+\\w+)*/[0-9]+(?:\\.[0-9]+)+=([^,]+)

Regex demo