Separating a string in java

**Aaliya Seth** · 10-02-2010

Hello,
I need to cut a string into several pieces but retaining the separator in pieces.
example:
my chain "or (varname) is a (Subject)
Expected Result: [ "be", "(varname)", "being one", "(Subject)]
separators are the brackets ().
I tried several solutions (split, StringTokenizer ...) but I can only have [ "be", "varname", "being one", "Subject"]
Someone would have an idea? Thanks in advance.

**cyber-noob** · 10-02-2010

Hello,
Just be your own separation method using indexOf indexOf ('{') and ('}'). You search the '(', if you think you are looking for the ')' nearest you make a substring and do it again for the rest of the chain. I think this should take long time to code such a program , it is quit simple. If you need help about the class and the methods provided by the java you can find then on the official site of the sun.

**HardWeaR** · 10-02-2010

Hello,
You can use regular expressions, but if I do not control enough to find the solution. I will still file the snippet I just tested:

Code:

String s = "This is (variable) (I) variable2 sing";
List l = findVars(chain);
System.out.System.out.println(l);

Here is the code

Code:

@ SuppressWarnings("unchecked")
Public static List findVars(String input) {
		Pattern p = Pattern.compiles("\\{.*\\}");
Matcher m = p.matcher(input);
List list = new ArrayList();
		if (m.find()) {
			do {
				list.add(m.group());
				
			} while (m.find());
			return list;
		}
		return null;
}

**SoftWore** · 10-02-2010

Hello,
This is not the solution to the problem in question, but just for info, the "*" RegExp is the default "greedy" (greedy), i.e that "{.*} in a string like "(variable) is my () variable2" will matcher whole chain, including the first ')' encountered. One can remedy this by adding "?" after the *, which gives {.*?} which allow to stop the very first ")" found.

**GlassFish** · 10-02-2010

Hello,
Here is the code that can help you, just go through it

Code:

	private List <String> spwtdel(String str, String OPENLiMiT, String closeLimit)
	{
		List lt = <String> new ArrayList <String>();
		int froplm = str.indexOf(OPENLiMiT);
		int frcllm = str.indexOf(closeLimit, froplm);
		while ((froplm! = -1) && (frcllm! = -1))
		{
			if (froplm> 0)
				lt.add(str.substring(0, FirstOpenLimit));
lt.add(str.substring(froplm, frcllm + 1));
			
			if (frcllm <str.length())
				str = str.substring(frcllm + 1, Str.length());
			else
				str = "";
froplm = str.indexOf(OPENLiMiT);
frcllm = str.indexOf(closeLimit, froplm);			
		}
		if (str.length() > 0)
			lt.add(str);
		return lt;
	}

**NetWorkInG** · 10-02-2010

Hello,
Here is another program that will interest you. Have a look at it

Code:

List <String> split(String string) {
        ArrayList str = <String> new ArrayList <String>();

Pattern pat = Pattern.compiles("\\{(.*?)\\}");
Matcher mat = pat.mat(string);
				
	int ls = 0;
	while (mat.find()) {
	  str.add(string.substring(ls, match.home()));
          str.add(mat.group());
ls = mat.end();
	}
	if (ls! = string.length())
	  str.add(string.substring(ls, string.length()));
 
         return str;
}