I have a multiline string which is delimited by a set of different delimiters:


I can split this string into its parts, using String.split, but it seems that I can't get the actual string, which matched the delimiter regex.

In other words, this is what I get:

  • Text1
  • Text2
  • Text3
  • Text4

This is what I want

  • Text1
  • DelimiterA
  • Text2
  • DelimiterC
  • Text3
  • DelimiterB
  • Text4

Is there any JDK way to split the string using a delimiter regex but also keep the delimiters?

Solution 1

You can use lookahead and lookbehind, which are features of regular expressions.


And you will get:

[a;, b;, c;, d]
[a, ;b, ;c, ;d]
[a, ;, b, ;, c, ;, d]

The last one is what you want.

((?<=;)|(?=;)) equals to select an empty character before ; or after ;.

EDIT: Fabian Steeg's comments on readability is valid. Readability is always a problem with regular expressions. One thing I do to make regular expressions more readable is to create a variable, the name of which represents what the regular expression does. You can even put placeholders (e.g. %1$s) and use Java's String.format to replace the placeholders with the actual string you need to use; for example:

static public final String WITH_DELIMITER = "((?<=%1$s)|(?=%1$s))";

public void someMethod() {
    final String[] aEach = "a;b;c;d".split(String.format(WITH_DELIMITER, ";"));

Solution 2

You want to use lookarounds, and split on zero-width matches. Here are some examples:

public class SplitNDump {
    static void dump(String[] arr) {
        for (String s : arr) {
            System.out.format("[%s]", s);
    public static void main(String[] args) {
        // "[1][234][567][890]"
        // "[1][,234][,567][,890]"
        // "[1,][234,][567,][890]"
        // "[1][,][234][,][567][,][890]"

        // "[][:][a][:][bb][:][:][c][:]"
        // "[:][a][:][bb][:][:][c][:]"
        dump(":::a::::b  b::c:".split("(?=(?!^):)(?<!:)|(?!:)(?<=:)"));
        // "[:::][a][::::][b  b][::][c][:]"
        dump("a,bb:::c  d..e".split("(?!^)\\b"));
        // "[a][,][bb][:::][c][  ][d][..][e]"

        // "[Array][Index][Out][Of][Bounds][Exception]"
        // "[1234][5678][90]"

        // Split at the end of each run of letter
        dump("Boooyaaaah! Yippieeee!!".split("(?<=(?=(.)\\1(?!\\1))..)"));
        // "[Booo][yaaaa][h! Yipp][ieeee][!!]"

And yes, that is triply-nested assertion there in the last pattern.

Related questions

See also

Solution 3

A very naive solution, that doesn't involve regex would be to perform a string replace on your delimiter along the lines of (assuming comma for delimiter):

string.replace(FullString, "," , "~,~")

Where you can replace tilda (~) with an appropriate unique delimiter.

Then if you do a split on your new delimiter then i believe you will get the desired result.

Solution 4

import java.util.regex.*;
import java.util.LinkedList;

public class Splitter {
    private static final Pattern DEFAULT_PATTERN = Pattern.compile("\\s+");

    private Pattern pattern;
    private boolean keep_delimiters;

    public Splitter(Pattern pattern, boolean keep_delimiters) {
        this.pattern = pattern;
        this.keep_delimiters = keep_delimiters;
    public Splitter(String pattern, boolean keep_delimiters) {
        this(Pattern.compile(pattern==null?"":pattern), keep_delimiters);
    public Splitter(Pattern pattern) { this(pattern, true); }
    public Splitter(String pattern) { this(pattern, true); }
    public Splitter(boolean keep_delimiters) { this(DEFAULT_PATTERN, keep_delimiters); }
    public Splitter() { this(DEFAULT_PATTERN); }

    public String[] split(String text) {
        if (text == null) {
            text = "";

        int last_match = 0;
        LinkedList<String> splitted = new LinkedList<String>();

        Matcher m = this.pattern.matcher(text);

        while (m.find()) {


            if (this.keep_delimiters) {

            last_match = m.end();


        return splitted.toArray(new String[splitted.size()]);

    public static void main(String[] argv) {
        if (argv.length != 2) {
            System.err.println("Syntax: java Splitter <pattern> <text>");

        Pattern pattern = null;
        try {
            pattern = Pattern.compile(argv[0]);
        catch (PatternSyntaxException e) {

        Splitter splitter = new Splitter(pattern);

        String text = argv[1];
        int counter = 1;
        for (String part : splitter.split(text)) {
            System.out.printf("Part %d: \"%s\"\n", counter++, part);

    > java Splitter "\W+" "Hello World!"
    Part 1: "Hello"
    Part 2: " "
    Part 3: "World"
    Part 4: "!"
    Part 5: ""

I don't really like the other way, where you get an empty element in front and back. A delimiter is usually not at the beginning or at the end of the string, thus you most often end up wasting two good array slots.

Edit: Fixed limit cases. Commented source with test cases can be found here: http://snippets.dzone.com/posts/show/6453

Solution 5

Pass the 3rd aurgument as "true". It will return delimiters as well.

StringTokenizer(String str, String delimiters, true);

Solution 6

I got here late, but returning to the original question, why not just use lookarounds?

Pattern p = Pattern.compile("(?<=\\w)(?=\\W)|(?<=\\W)(?=\\w)");


[', ab, ',', cd, ',', eg, ']
[boo, :, and, :, foo]

EDIT: What you see above is what appears on the command line when I run that code, but I now see that it's a bit confusing. It's difficult to keep track of which commas are part of the result and which were added by Arrays.toString(). SO's syntax highlighting isn't helping either. In hopes of getting the highlighting to work with me instead of against me, here's how those arrays would look it I were declaring them in source code:

{ "'", "ab", "','", "cd", "','", "eg", "'" }
{ "boo", ":", "and", ":", "foo" }

I hope that's easier to read. Thanks for the heads-up, @finnw.

Solution 7

I know this is a very-very old question and answer has also been accepted. But still I would like to submit a very simple answer to original question. Consider this code:

String str = "Hello-World:How\nAre You&doing";
inputs = str.split("(?!^)\\b");
for (int i=0; i<inputs.length; i++) {
   System.out.println("a[" + i + "] = \"" + inputs[i] + '"');


a[0] = "Hello"
a[1] = "-"
a[2] = "World"
a[3] = ":"
a[4] = "How"
a[5] = "
a[6] = "Are"
a[7] = " "
a[8] = "You"
a[9] = "&"
a[10] = "doing"

I am just using word boundary \b to delimit the words except when it is start of text.

Solution 8

I had a look at the above answers and honestly none of them I find satisfactory. What you want to do is essentially mimic the Perl split functionality. Why Java doesn't allow this and have a join() method somewhere is beyond me but I digress. You don't even need a class for this really. Its just a function. Run this sample program:

Some of the earlier answers have excessive null-checking, which I recently wrote a response to a question here:


Anyway, the code:

public class Split {
    public static List<String> split(String s, String pattern) {
        assert s != null;
        assert pattern != null;
        return split(s, Pattern.compile(pattern));

    public static List<String> split(String s, Pattern pattern) {
        assert s != null;
        assert pattern != null;
        Matcher m = pattern.matcher(s);
        List<String> ret = new ArrayList<String>();
        int start = 0;
        while (m.find()) {
            ret.add(s.substring(start, m.start()));
            start = m.end();
        ret.add(start >= s.length() ? "" : s.substring(start));
        return ret;

    private static void testSplit(String s, String pattern) {
        System.out.printf("Splitting '%s' with pattern '%s'%n", s, pattern);
        List<String> tokens = split(s, pattern);
        System.out.printf("Found %d matches%n", tokens.size());
        int i = 0;
        for (String token : tokens) {
            System.out.printf("  %d/%d: '%s'%n", ++i, tokens.size(), token);

    public static void main(String args[]) {
        testSplit("abcdefghij", "z"); // "abcdefghij"
        testSplit("abcdefghij", "f"); // "abcde", "f", "ghi"
        testSplit("abcdefghij", "j"); // "abcdefghi", "j", ""
        testSplit("abcdefghij", "a"); // "", "a", "bcdefghij"
        testSplit("abcdefghij", "[bdfh]"); // "a", "b", "c", "d", "e", "f", "g", "h", "ij"

Solution 9

I like the idea of StringTokenizer because it is Enumerable.
But it is also obsolete, and replace by String.split which return a boring String[] (and does not includes the delimiters).

So I implemented a StringTokenizerEx which is an Iterable, and which takes a true regexp to split a string.

A true regexp means it is not a 'Character sequence' repeated to form the delimiter:
'o' will only match 'o', and split 'ooo' into three delimiter, with two empty string inside:

[o], '', [o], '', [o]

But the regexp o+ will return the expected result when splitting "aooob"

[], 'a', [ooo], 'b', []

To use this StringTokenizerEx:

final StringTokenizerEx aStringTokenizerEx = new StringTokenizerEx("boo:and:foo", "o+");
final String firstDelimiter = aStringTokenizerEx.getDelimiter();
for(String aString: aStringTokenizerEx )
    // uses the split String detected and memorized in 'aString'
    final nextDelimiter = aStringTokenizerEx.getDelimiter();

The code of this class is available at DZone Snippets.

As usual for a code-challenge response (one self-contained class with test cases included), copy-paste it (in a 'src/test' directory) and run it. Its main() method illustrates the different usages.

Note: (late 2009 edit)

The article Final Thoughts: Java Puzzler: Splitting Hairs does a good work explaning the bizarre behavior in String.split().
Josh Bloch even commented in response to that article:

Yes, this is a pain. FWIW, it was done for a very good reason: compatibility with Perl.
The guy who did it is Mike "madbot" McCloskey, who now works with us at Google. Mike made sure that Java's regular expressions passed virtually every one of the 30K Perl regular expression tests (and ran faster).

The Google common-library Guava contains also a Splitter which is:

  • simpler to use
  • maintained by Google (and not by you)

So it may worth being checked out. From their initial rough documentation (pdf):

JDK has this:

String[] pieces = "foo.bar".split("\\.");

It's fine to use this if you want exactly what it does: - regular expression - result as an array - its way of handling empty pieces

Mini-puzzler: ",a,,b,".split(",") returns...

(a) "", "a", "", "b", ""
(b) null, "a", null, "b", null
(c) "a", null, "b"
(d) "a", "b"
(e) None of the above

Answer: (e) None of the above.

"", "a", "", "b"

Only trailing empties are skipped! (Who knows the workaround to prevent the skipping? It's a fun one...)

In any case, our Splitter is simply more flexible: The default behavior is simplistic:

Splitter.on(',').split(" foo, ,bar, quux,")
--> [" foo", " ", "bar", " quux", ""]

If you want extra features, ask for them!

.split(" foo, ,bar, quux,")
--> ["foo", "bar", "quux"]

Order of config methods doesn't matter -- during splitting, trimming happens before checking for empties.

Solution 10

Here is a simple clean implementation which is consistent with Pattern#split and works with variable length patterns, which look behind cannot support, and it is easier to use. It is similar to the solution provided by @cletus.

public static String[] split(CharSequence input, String pattern) {
    return split(input, Pattern.compile(pattern));

public static String[] split(CharSequence input, Pattern pattern) {
    Matcher matcher = pattern.matcher(input);
    int start = 0;
    List<String> result = new ArrayList<>();
    while (matcher.find()) {
        result.add(input.subSequence(start, matcher.start()).toString());
        start = matcher.end();
    if (start != input.length()) result.add(input.subSequence(start, input.length()).toString());
    return result.toArray(new String[0]);

I don't do null checks here, Pattern#split doesn't, why should I. I don't like the if at the end but it is required for consistency with the Pattern#split . Otherwise I would unconditionally append, resulting in an empty string as the last element of the result if the input string ends with the pattern.

I convert to String[] for consistency with Pattern#split, I use new String[0] rather than new String[result.size()], see here for why.

Here are my tests:

public void splitsVariableLengthPattern() {
    String[] result = Split.split("/foo/$bar/bas", "\\$\\w+");
    Assert.assertArrayEquals(new String[] { "/foo/", "$bar", "/bas" }, result);

public void splitsEndingWithPattern() {
    String[] result = Split.split("/foo/$bar", "\\$\\w+");
    Assert.assertArrayEquals(new String[] { "/foo/", "$bar" }, result);

public void splitsStartingWithPattern() {
    String[] result = Split.split("$foo/bar", "\\$\\w+");
    Assert.assertArrayEquals(new String[] { "", "$foo", "/bar" }, result);

public void splitsNoMatchesPattern() {
    String[] result = Split.split("/foo/bar", "\\$\\w+");
    Assert.assertArrayEquals(new String[] { "/foo/bar" }, result);

Solution 11

I will post my working versions also(first is really similar to Markus).

public static String[] splitIncludeDelimeter(String regex, String text){
    List<String> list = new LinkedList<>();
    Matcher matcher = Pattern.compile(regex).matcher(text);

    int now, old = 0;
        now = matcher.end();
        list.add(text.substring(old, now));
        old = now;

    if(list.size() == 0)
        return new String[]{text};

    //adding rest of a text as last element
    String finalElement = text.substring(old);

    return list.toArray(new String[list.size()]);

And here is second solution and its round 50% faster than first one:

public static String[] splitIncludeDelimeter2(String regex, String text){
    List<String> list = new LinkedList<>();
    Matcher matcher = Pattern.compile(regex).matcher(text);

    StringBuffer stringBuffer = new StringBuffer();
        matcher.appendReplacement(stringBuffer, matcher.group());
        stringBuffer.setLength(0); //clear buffer

    matcher.appendTail(stringBuffer); ///dodajemy reszte  ciagu

    return list.toArray(new String[list.size()]);

Solution 12

Another candidate solution using a regex. Retains token order, correctly matches multiple tokens of the same type in a row. The downside is that the regex is kind of nasty.

package javaapplication2;

import java.util.ArrayList;
import java.util.List;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class JavaApplication2 {

     * @param args the command line arguments
    public static void main(String[] args) {
        String num = "58.5+variable-+98*78/96+a/78.7-3443*12-3";

        // Terrifying regex:
        //  (a)|(b)|(c) match a or b or c
        // where
        //   (a) is one or more digits optionally followed by a decimal point
        //       followed by one or more digits: (\d+(\.\d+)?)
        //   (b) is one of the set + * / - occurring once: ([+*/-])
        //   (c) is a sequence of one or more lowercase latin letter: ([a-z]+)
        Pattern tokenPattern = Pattern.compile("(\\d+(\\.\\d+)?)|([+*/-])|([a-z]+)");
        Matcher tokenMatcher = tokenPattern.matcher(num);

        List<String> tokens = new ArrayList<>();

        while (!tokenMatcher.hitEnd()) {
            if (tokenMatcher.find()) {
            } else {
                // report error


Sample output:

[58.5, +, variable, -, +, 98, *, 78, /, 96, +, a, /, 78.7, -, 3443, *, 12, -, 3]

Solution 13

I don't know of an existing function in the Java API that does this (which is not to say it doesn't exist), but here's my own implementation (one or more delimiters will be returned as a single token; if you want each delimiter to be returned as a separate token, it will need a bit of adaptation):

static String[] splitWithDelimiters(String s) {
    if (s == null || s.length() == 0) {
        return new String[0];
    LinkedList<String> result = new LinkedList<String>();
    StringBuilder sb = null;
    boolean wasLetterOrDigit = !Character.isLetterOrDigit(s.charAt(0));
    for (char c : s.toCharArray()) {
        if (Character.isLetterOrDigit(c) ^ wasLetterOrDigit) {
            if (sb != null) {
            sb = new StringBuilder();
            wasLetterOrDigit = !wasLetterOrDigit;
    return result.toArray(new String[0]);

Solution 14

I suggest using Pattern and Matcher, which will almost certainly achieve what you want. Your regular expression will need to be somewhat more complicated than what you are using in String.split.

Solution 15

I don't think it is possible with String#split, but you can use a StringTokenizer, though that won't allow you to define your delimiter as a regex, but only as a class of single-digit characters:

new StringTokenizer("Hello, world. Hi!", ",.!", true); // true for returnDelims

Solution 16

If you can afford, use Java's replace(CharSequence target, CharSequence replacement) method and fill in another delimiter to split with. Example: I want to split the string "boo:and:foo" and keep ':' at its righthand String.

String str = "boo:and:foo";
str = str.replace(":","newdelimiter:");
String[] tokens = str.split("newdelimiter");

Important note: This only works if you have no further "newdelimiter" in your String! Thus, it is not a general solution. But if you know a CharSequence of which you can be sure that it will never appear in the String, this is a very simple solution.

Solution 17

Fast answer: use non physical bounds like \b to split. I will try and experiment to see if it works (used that in PHP and JS).

It is possible, and kind of work, but might split too much. Actually, it depends on the string you want to split and the result you need. Give more details, we will help you better.

Another way is to do your own split, capturing the delimiter (supposing it is variable) and adding it afterward to the result.

My quick test:

String str = "'ab','cd','eg'";
String[] stra = str.split("\\b");
for (String s : stra) System.out.print(s + "|");



A bit too much... :-)

Solution 18

Tweaked Pattern.split() to include matched pattern to the list


// add match to the list
        matchList.add(input.subSequence(start, end).toString());

Full source

public static String[] inclusiveSplit(String input, String re, int limit) {
    int index = 0;
    boolean matchLimited = limit > 0;
    ArrayList<String> matchList = new ArrayList<String>();

    Pattern pattern = Pattern.compile(re);
    Matcher m = pattern.matcher(input);

    // Add segments before each match found
    while (m.find()) {
        int end = m.end();
        if (!matchLimited || matchList.size() < limit - 1) {
            int start = m.start();
            String match = input.subSequence(index, start).toString();
            // add match to the list
            matchList.add(input.subSequence(start, end).toString());
            index = end;
        } else if (matchList.size() == limit - 1) { // last one
            String match = input.subSequence(index, input.length())
            index = end;

    // If no match was found, return this
    if (index == 0)
        return new String[] { input.toString() };

    // Add remaining segment
    if (!matchLimited || matchList.size() < limit)
        matchList.add(input.subSequence(index, input.length()).toString());

    // Construct result
    int resultSize = matchList.size();
    if (limit == 0)
        while (resultSize > 0 && matchList.get(resultSize - 1).equals(""))
    String[] result = new String[resultSize];
    return matchList.subList(0, resultSize).toArray(result);

Solution 19

Here's a groovy version based on some of the code above, in case it helps. It's short, anyway. Conditionally includes the head and tail (if they are not empty). The last part is a demo/test case.

List splitWithTokens(str, pat) {
    def tokens=[]
    def lastMatch=0
    def m = str=~pat
    while (m.find()) {
      if (m.start() > 0) tokens << str[lastMatch..<m.start()]
      tokens << m.group()
    if (lastMatch < str.length()) tokens << str[lastMatch..<str.length()]

[['<html><head><title>this is the title</title></head>',/<[^>]+>/],
 ['before<html><head><title>this is the title</title></head>after',/<[^>]+>/]
].each { 
   println splitWithTokens(*it)

Solution 20

An extremely naive and inefficient solution which works nevertheless.Use split twice on the string and then concatenate the two arrays

String temp[]=str.split("\\W");
String temp2[]=str.split("\\w||\\s");
int i=0;
for(String string:temp)
String temp3[]=new String[temp.length-1];
for(String string:temp2)
//      System.out.println(temp.length);
//      System.out.println(temp2.length);
String[] temp4=new String[temp.length+temp3.length];
int j=0;
for(String s:temp4)

Solution 21

    String expression = "((A+B)*C-D)*E";
    expression = expression.replaceAll("\\+", "~+~");
    expression = expression.replaceAll("\\*", "~*~");
    expression = expression.replaceAll("-", "~-~");
    expression = expression.replaceAll("/+", "~/~");
    expression = expression.replaceAll("\\(", "~(~"); //also you can use [(] instead of \\(
    expression = expression.replaceAll("\\)", "~)~"); //also you can use [)] instead of \\)
    expression = expression.replaceAll("~~", "~");
    if(expression.startsWith("~")) {
        expression = expression.substring(1);

    String[] expressionArray = expression.split("~");

Solution 22

One of the subtleties in this question involves the "leading delimiter" question: if you are going to have a combined array of tokens and delimiters you have to know whether it starts with a token or a delimiter. You could of course just assume that a leading delim should be discarded but this seems an unjustified assumption. You might also want to know whether you have a trailing delim or not. This sets two boolean flags accordingly.

Written in Groovy but a Java version should be fairly obvious:

            String tokenRegex = /[\p{L}\p{N}]+/ // a String in Groovy, Unicode alphanumeric
            def finder = phraseForTokenising =~ tokenRegex
            // NB in Groovy the variable 'finder' is then of class java.util.regex.Matcher
            def finderIt = finder.iterator() // extra method added to Matcher by Groovy magic
            int start = 0
            boolean leadingDelim, trailingDelim
            def combinedTokensAndDelims = [] // create an array in Groovy

            while( finderIt.hasNext() )
                def token = finderIt.next()
                int finderStart = finder.start()
                String delim = phraseForTokenising[ start  .. finderStart - 1 ]
                // Groovy: above gets slice of String/array
                if( start == 0 ) leadingDelim = finderStart != 0
                if( start > 0 || leadingDelim ) combinedTokensAndDelims << delim
                combinedTokensAndDelims << token // add element to end of array
                start = finder.end()
            // start == 0 indicates no tokens found
            if( start > 0 ) {
                // finish by seeing whether there is a trailing delim
                trailingDelim = start < phraseForTokenising.length()
                if( trailingDelim ) combinedTokensAndDelims << phraseForTokenising[ start .. -1 ]

                println( "leading delim? $leadingDelim, trailing delim? $trailingDelim, combined array:\n $combinedTokensAndDelims" )


Solution 23

If you want keep character then use split method with loophole in .split() method.

See this example:

public class SplitExample {

    public static void main(String[] args) {  
        String str = "Javathomettt";  
        System.out.println("method 1");
        System.out.println("Returning words:");  
        String[] arr = str.split("t", 40);  
        for (String w : arr) {  
        System.out.println("Split array length: "+arr.length);  
        System.out.println("method 2");
        System.out.println(str.replaceAll("t", "\n"+"t"));

Solution 24

I don't know Java too well, but if you can't find a Split method that does that, I suggest you just make your own.

string[] mySplit(string s,string delimiter)
    string[] result = s.Split(delimiter);
    for(int i=0;i<result.Length-1;i++)
        result[i] += delimiter; //this one would add the delimiter to each items end except the last item, 
                    //you can modify it however you want
string[] res = mySplit(myString,myDelimiter);

Its not too elegant, but it'll do.