如何创建正则表达式来替换已知字符串，同时保持可选参数不变？

Question

I've been trying out now for a while, but don't get it right: In Java, I am trying to create a regular expression to match and replace a (to me known) string out of a string while keeping optional parameters intact.

Example inputs:

{067e6162-3b6f-4ae2-a171-2470b63dff00}
{067e6162-3b6f-4ae2-a171-2470b63dff00,number}
{067e6162-3b6f-4ae2-a171-2470b63dff00,number,integer}
{067e6162-3b6f-4ae2-a171-2470b63dff00,choice,1#one more item|1<another {067e6162-3b6f-4ae2-a171-2470b63dff00,number,integer} items}

(Note that the last example contains a nested reference to the same input string). The format is always enclosing the to-be-replaced string in curly brackets {...} but with an optional list of comma-separated parameter(s).

I want to replace the input string with a number, eg for above input strings the result should be:

{2}
{2,number}
{2,number,integer}
{2,choice,1#one more item|1<another {2,number,integer} items}

Ideally, I'd like to have a regex that is flexible enough to handle (almost) any string as pattern to be replaced, so not just UUID kind of strings as above but also something like this:

A test string with {the_known_input_value_to_be_replaced,number,integer} not replacing the_known_input_value_to_be_replaced if its not in curly brackets of course.

which should end up as eg:

A test string with {3,number,integer} not replacing the_known_input_value_to_be_replaced if its not in curly brackets of course.

Note that the substitution should only take place if the input string is in curly brackets. In Java I will be able to construct the pattern at runtime, taking the to-be-replaced string into account verbosely.

I tried eg \\{(067e6162-3b6f-4ae2-a171-2470b63dff00)(,?.*)\\} (not java escaped yet) and more generic approaches like \\{(+?)(,?.*)\\} , but they all don't do it right.

Any advice from regex ninjas highly appreciated :)

Answer 1

If you the known old string always occurs right after { you can just use

String result = old_text.replace("{" + my_old_keyword, "{" + my_new_keyword);

If you really have multiple known strings inside curly brackets (and there are no escaped curly brackets to take care of), you can use the following code:

String input = "067e6162-3b6f-4ae2-a171-2470b63dff00 is outside {067e6162-3b6f-4ae2-a171-2470b63dff00,choice,067e6162-3b6f-4ae2-a171-2470b63dff00,1#one more item|1<another {067e6162-3b6f-4ae2-a171-2470b63dff00,number,067e6162-3b6f-4ae2-a171-2470b63dff00,integer} items} 067e6162-3b6f-4ae2-a171-2470b63dff00 is outside ";
String old_key = "067e6162-3b6f-4ae2-a171-2470b63dff00";
String new_key = "NEW_KEY";
List<String> chunks = replaceInBalancedSubstrings(input, '{', '}', old_key, new_key);
System.out.println(String.join("", chunks));

Result: 067e6162-3b6f-4ae2-a171-2470b63dff00 is outside {{NEW_KEY,choice,NEW_KEY,1#one more item|1<another {NEW_KEY,number,NEW_KEY,integer} items} 067e6162-3b6f-4ae2-a171-2470b63dff00 is outside

The replaceInBalancedSubstrings method will look like:

public static List<String> replaceInBalancedSubstrings(String s, Character markStart, Character markEnd, String old_key, String new_key) {
    List<String> subTreeList = new ArrayList<String>();
    int level = 0;
    int prevStart = 0;
    StringBuffer sb = new StringBuffer();
    int lastOpenBracket = -1;
    for (int i = 0; i < s.length(); i++) {
        char c = s.charAt(i);
        if (level == 0) {
            sb.append(c);
        }
        if (c == markStart) {
            level++;
            if (level == 1) {
                lastOpenBracket = i;
                if (sb.length() > 0) {
                    subTreeList.add(sb.toString());
                    sb.delete(0, sb.length());
                }
            }
        }
        else if (c == markEnd) {
            if (level == 1) {
                subTreeList.add(s.substring(lastOpenBracket, i+1).replace(old_key, new_key)); // String replacement here
            }
            level--;
        }
    }
    if (sb.length() > 0) {
        subTreeList.add(sb.toString());
    }
    return subTreeList;
}

See IDEONE demo

This code will deal with replacements only inside substrings inside balanced (nested) curly braces.