Java, Regex: Replace fields in Json by field key

Question

Requirements:

For each Json field where key matches specified constant replace value with another constant.

{"regular":"a", "sensitive":"b"}

Parameters "sensitive", "*****". Expected:

{"regular":"a", "sensitive":"*****"}

Values may, or may not have double quotes around them. Replacement constant is double quouted always. Json may be malformed. Java implementation preferably.

Key comparison is case insensitive.

Not a good idea to manipulate JSON using regex. Use some JSON library — TheLostMind
– TheLostMind, Commented Mar 7, 2018 at 7:19
@TheLostMind I have a requirement that Json may be malformed. — Cortlendt
– Cortlendt, Commented Mar 7, 2018 at 7:22
hmm. Well, in that case, there are ways to achieve this. Even though I've added an answer to this, I strongly recommend avoiding it — TheLostMind
– TheLostMind, Commented Mar 7, 2018 at 7:33

Tim Pietzcker · Accepted Answer · 2018-03-07 07:34:50Z

Depending on how malformed your "JSON" is, the following might work - if not, we need more test cases:

"sensitive"\s*:\s*  # match "sensitive":
(                   # capture in group 1:
 "[^"]*"            # any quoted value
|                   # or
 [^\s,{}"]*         # any unquoted value, ending at a comma, brace or whitespace
)                   # end of group 1

In Java:

String resultString = subjectString.replaceAll(
    "(?x)\"sensitive\"\\s*:\\s* # match \"sensitive\":\n" +
    "(                          # capture in group 1:\n" +
    " \"[^\"]*\"                # any quoted value\n" +
    "|                          # or\n" +
    " [^\\s,{}\"]*              # an unquoted value, ending at comma, brace or whitespace\n" +
    ")                          # end of group 1",
    "\"sensitive\":\"******\"");

Test it live on regex101.com.

TheLostMind · Accepted Answer · 2018-03-07 07:32:21Z

3

You can use positive lookbehind to achieve this :

public static void main(String[] args) {
    String s = "{\"regular\":\"a\", \"sensitive\":\"b\"}";
    String key = "sensitive";
    String val = "****";
    System.out.println(s.replaceAll("(?<=\"" + key + "\":\")(\\w+)", val));
     key = "regular";
    System.out.println(s.replaceAll("(?<=\"" + key + "\":\")(\\w+)", val));

}

O/P :

{"regular":"a", "sensitive":"****"}
{"regular":"****", "sensitive":"b"}

answered Mar 7, 2018 at 7:32

TheLostMind

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3 Comments

Cortlendt Over a year ago

how can I make key comparison case-insenstive?

Cortlendt Over a year ago

used (?i) for case insensitive matching

Robel Robel Lingstuyl Over a year ago

what would you do if "sensitive" were an object? {sensitive:{x:y} and you wanted to wipe the object and put in the string?

Allan · Accepted Answer · 2018-03-07 07:37:07Z

1

You can use the following regex:

String t= "{\"regular\":\"a\", \"sensitive\":\"b\"}"; //{"regular":"a", "sensitive":"b"}
String r = t.replaceAll("(\\s*)\"?sensitive\"?\\s*:\\s*\"?b\"?\\s*", "$1\"sensitive\":\"*****\""); 
System.out.println("output "+r); //output {"regular":"a", "sensitive":"*****"}

t= "{\"regular\":\"a\",sensitive:b}"; //{"regular":"a", "sensitive":"b"}
r = t.replaceAll("(\\s*)\"?sensitive\"?\\s*:\\s*\"?b\"?\\s*", "$1\"sensitive\":\"*****\""); 
System.out.println("output "+r); //output {"regular":"a","sensitive":"*****"}

DEMO: https://regex101.com/r/uHUhEl/1/

answered Mar 7, 2018 at 7:37

Allan

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