Pattern used with split

[Aaliya Seth]

Hello,
I wonder if it is possible to maintain the pattern used with the split method to split a sentence.
For example with

String sen = "The little cat."
String[] phr = sen.split("small");
/ / phr [0] contains "The" and phr [1] contains "cat."

Is it possible to have some work such phr [0] = "The" then phr[1] = "small" (pattern cutting) and phr [2] = "cat". Any help on this will be highly appreciated. Thanks in advance.

[cyber-noob]

Hello,
Yes it is possible, but using regular expressions instead of a split:

String sentence = "kitten";
Pattern pat = Pattern.compiles("^(.+)( small )(.+)$");
Matcher mat = pat.matcher(sentence);
 
if(mat.games()) {
   System.out.System.out.println(mat.group(1)); / / "My"
   System.out.System.out.println(mat.group(2)); / / "Small"
   System.out.System.out.println(mat.group(3)); / / Chat
}

Try this and see if the code works perfect for you.

[HardWeaR]

Hello,
I think instead of the split with "small", you do it with one or more spaces:
Here is the code for it

String[] phr = sentence.split(" +");

I do not know what you exactly want the output of your code, but if you need this as the output then it will be a great option for your program. Try it once and if not successful, then leave it and try some other option.

[InterNetWorKed]

Hello,
In my opinion no, no such base. It remains you to retrace the table later to create a second table with your separators (here "small") between your words. Split with the "+" does not answer the question in my mind as to the specific example given. The regular expression not because it would require to know in advance how much of a "small" in the chain. Hope you are understanding the circumstances of it.

[WinDoWLoCuS]

Hello,
You can try the following code

Public static String[] split(String in, String reg, boolean incsep) {
		/ / We create a matcher that will search all partitions:
		Matcher m = Pattern.compiles(reg).matcher(in);
 
		/ / List of 'words' to return:
		List <String> mtlst = new ArrayList <String>();
		/ / Index out of words to be cut:
		int index = 0;
 
		/ / It searches for each separator:
		while(m.find()) {
			/ / We cut the string before the separator found:
			mtlst.add( in.substring(index, m.home()) );
			
			if (incsep) {
				/ / It returns the separator:
				mtlst.add( m.group() );
			}
			
			/ / We move the index after the separator
			index = m.end();
		}
		
		if (Index ==0) {
			/ / Special case: no separator found
			/ / It returns the entire string:
			return new String[] {in};
		}
		
		if (index <in.length()) {
			/ / We add the end of the chain (last word)
			mtlst.add( in.substring(Index) );
		}
		return mtlst.toArray(new String[mtlst.size()]);
	}

[Linux-Us]

Hell,
Eclipse, when you have this error message, simply go left into the small cross of the error. Among the solutions to resolve the error, he offers hopefully "Change source level to ". This is the solution to be learned. Otherwise, the alternative is to make the code compatible java 1.4 (which is simply to remove "<String>). If you need more information on Eclipse the you can visit their official site.