如何获取java中源字符串中大括号内的字符串数组

Question

There is a string object in java with its contents as:

String sourceString="This {is} a sample {string} that {contains} {substrings} inside curly {braces}";

I want the array of string with its contents as: {is},{string},{contains},{substrings}{braces}

Following is the code that I wrote to get the result but the output I am getting is:

"{is} a sample {string} that {contains} {substrings} inside curly {braces}"

So, basically it is taking all the characters that are in between the first open curly braces and last closing curly braces.

// Source string
String sourceString="This {is} a samle {string} that {contains} {substrings} inside curly {braces}";

// Regular expression to get the values between curly braces (there is a mistake, I guess)
String regex="\\{(.*)\\}";
Matcher matcher = Pattern.compile(regex).matcher(sourceString);

while (matcher.find()) {
    System.out.println(matcher.group(0));
}

Answer 1

A little bit of Googling found this solution, which gave me ideas for the pattern

Some time spent with Lesson: Regular Expressions brought the libraries and functionality I would need to provide this example...

String exp = "\\{(.*?)\\}";

String value = "This {is} a samle {string} that {contains} {substrings} inside curly {braces}";

Pattern pattern = Pattern.compile(exp);
Matcher matcher = pattern.matcher(value);

List<String> matches = new ArrayList<String>(5);
while (matcher.find()) {
    String group = matcher.group();
    matches.add(group);
}

String[] groups = matches.toArray(new String[matches.size()]);
System.out.println(Arrays.toString(groups));

Which outputs

[{is}, {string}, {contains}, {substrings}, {braces}]

Answer 2

Here's the one line solution:

String[] resultArray = str.replaceAll("^[^{]*|[^}]*$", "").split("(?<=\\})[^{]*");

This works by first stripping off the leading and trailing junk, then splitting on everything between } and { .

---

Here's some test code:

String str = "This {is} a samle {string} that {contains} {substrings} inside curly";
String[] resultArray = str.replaceAll("^[^{]*|[^}]*$", "").split("(?<=\\})[^{]*");
System.out.println(Arrays.toString(resultArray));

Output:

[{is}, {string}, {contains}, {substrings}]

Answer 3

• Pattern which will match {characters} can look like \\\\{[^}]*\\\\} .
• Now using Pattern and Matcher classes you can find each substring that matches this regex.
• Place each of founded substring in List<String>
• After list is filled with all substrings you can convert it to array using yourList.toArray(newStringArray) method.

---

EDIT after your update

Problem with your regex is that * quantifier is greedy which means it will try to find maximal possible solution. So in case of \\\\{(.*)\\\\} it will match

• first possible { ,
• zero or more characters

• last possible } which in case of

   This {is} a samle {string} that {contains} {substrings} inside curly {braces} 

means it will start from { in {is} and finis in } from {braces}

To make * find minimal set of characters which can be used to create matching substring you need to either

• add ? after it making *? quantifier reluctant,
• describe your regex as I did originally and exclude } from possible match between { and } , so instead of matching any characters which represents . use [^}] which represents any character except } .