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How can I make the following regex match my censors? Java

I am trying to censor specific strings, and patterns within my application but my matcher doesn't seem to be finding any results when searching for the Pattern.
public String censorString(String s) {
System.out.println("Censoring... "+ s);
if (findPatterns(s)) {
System.out.println("Found pattern");
for (String censor : foundPatterns) {
for (int i = 0; i < censor.length(); i++)
s.replace(censor.charAt(i), (char)42);
}
}
return s;
}
public boolean findPatterns(String s) {
for (String censor : censoredWords) {
Pattern p = Pattern.compile("(.*)["+censor+"](.*)");//regex
Matcher m = p.matcher(s);
while (m.find()) {
foundPatterns.add(censor);
return true;
}
}
return false;
}
At the moment I'm focusing on just the one pattern, if the censor is found in the string. I've tried many combinations and none of them seem to return "true".
"(.*)["+censor+"](.*)"
"(.*)["+censor+"]"
"["+censor+"]"
"["+censor+"]+"
Any help would be appreciated.
Usage: My censored words are "hello", "goodbye"
String s = "hello there, today is a fine day."
System.out.println(censorString(s));
is supposed to print " ***** today is a fine day. "

Your regex is right!!!!. The problem is here.
s.replace(censor.charAt(i), (char)42);
If you expect this line to rewrite the censored parts of your string it will not. Please check the java doc for string.
Please find below the program which will do what you intend to do. I removed your findpattern method and just used the replaceall with regex in String API. Hope this helps.
public class Regex_SO {
private String[] censoredWords = new String[]{"hello"};
/**
* #param args the command line arguments
*/
public static void main(String[] args) {
Regex_SO regex_SO = new Regex_SO();
regex_SO.censorString("hello there, today is a fine day. hello again");
}
public String censorString(String s) {
System.out.println("Censoring... "+ s);
for(String censoredWord : censoredWords){
String replaceStr = "";
for(int index = 0; index < censoredWord.length();index++){
replaceStr = replaceStr + "*";
}
s = s.replaceAll(censoredWord, replaceStr);
}
System.out.println("Censored String is .. " + s);
return s;
}
}

Since this seem like homework I cant give you working code, but here are few pointers
consider using \\b(word1|word2|word3)\\b regex to find specific words
to create char representing * you can write it as '*'. Don't use (char)42 to avoid magic numbers
to create new string which will have same length as old string but will be filled with only specific characters you can use String newString = oldString.replaceAll(".","*")
to replace on-the-fly founded match with new value you can use appendReplacement and appendTail methods from Matcher class. Here is how code using it should look like
StringBuffer sb = new StringBuffer();//buffer for string with replaced values
Pattern p = Pattern.compile(yourRegex);
Matcher m = p.matcher(yourText);
while (m.find()){
String match = m.group(); //this will represent current match
String newValue = ...; //here you need to decide how to replace it
m.appentReplacemenet(sb, newValue );
}
m.appendTail(sb);
String censoredString = sb.toString();

how to use one string to match many rules？

Conditions:
there are many rules ,maybe hundreds, which are like :
{aab*, aabc*,
aabcdd*, dtctddds*,
*ddt*,
*cddt*,
*bcddt*,
*t,
*ttt,
*ccddttt}
each time I will get one string, then I should find the longest matched rule.
Examples:
example 1.string is aabcddttt the matched rule should be: aabcdd*
example 2. string is accddttt the matched rule should be *ccddttt
Question:
I don't want to use the rules in a long array to match the string one by one,that is inefficient method.maybe I should use the string as a regex to match the hundred rules.But yet I can't find a elegant way to solve this problem.
Can I use some regexes to get the result?
Which is the best/fastest way to match?
Java, plain C or shell are preferred,please don't use C++ STL

Longest common substring
Perhaps this algorithm is what you are looking for =).
Why not do it simply?
String[] rules = {"^aab", "bcd", "aabcdd$", "dtctddds$", "^ddt$", "^cddt$", "^bcddt$", "^t", "^ttt", "^ccddttt"};
String testCase = "aabcddttt";
for (int i = 0; i < rules.length; i++) {
Pattern p = Pattern.compile(rules[i]);
Matcher m = p.matcher(testCase);
if (m.find()) {
System.out.println("String: " + testCase + " has matched the pattern " + rules[i]);
}
}
So basically in this case, rules[0], which is ^aab found because carrot (^) means string must begin with ^aab. On the other hand, bba$ means string must end with bba. And rules1 is found because it means the rule can appear anywhere from the testCase (e.g. bcd).

You could try matching them all at once with a brackets around each sub-rule. You could use the group to determine which matched.
public static void main(String... ignored) {
for (String test : "aabaa,wwwaabcdddd,abcddtxyz".split(",")) {
System.out.println(test + " matches " + longestMatch(test, "aab*", "aabc*", "aabcdd*", "dtctddds*", "ddt"));
}
}
public static String longestMatch(String text, String... regex) {
String[] sortedRegex = regex.clone();
Arrays.sort(sortedRegex, new Comparator<String>() {
#Override
public int compare(String o1, String o2) {
return o2.length() - o1.length();
}
});
StringBuilder sb = new StringBuilder();
String sep = "(";
for (String s : sortedRegex) {
sb.append(sep).append('(').append(s).append(')');
sep = "|";
}
sb.append(")");
Matcher matcher = Pattern.compile(sb.toString()).matcher(text);
if (matcher.find()) {
for (int i = 2; i <= matcher.groupCount(); i++) {
String group = matcher.group(i);
if (group != null)
return sortedRegex[i - 2];
}
}
return "";
}
prints
aabaa matches aabc*
wwwaabcdddd matches aabcdd*
abcddtxyz matches ddt

Performing multiple string replacements with metacharacter regex patterns

I am trying to perform multiple string replacements using Java's Pattern and Matcher, where the regex pattern may include metacharacters (e.g. \b, (), etc.). For example, for the input string fit i am, I would like to apply the replacements:
\bi\b --> EYE
i --> I
I then followed the coding pattern from two questions (Java Replacing multiple different substring in a string at once, Replacing multiple substrings in Java when replacement text overlaps search text). In both, they create an or'ed search pattern (e.g foo|bar) and a Map of (pattern, replacement), and inside the matcher.find() loop, they look up and apply the replacement.
The problem I am having is that the matcher.group() function does not contain information on matching metacharacters, so I cannot distinguish between i and \bi\b. Please see the code below. What can I do to fix the problem?
import java.util.regex.Matcher;
import java.util.regex.Pattern;
import java.util.*;
public class ReplacementExample
{
public static void main(String argv[])
{
Map<String, String> replacements = new HashMap<String, String>();
replacements.put("\\bi\\b", "EYE");
replacements.put("i", "I");
String input = "fit i am";
String result = doit(input, replacements);
System.out.printf("%s\n", result);
}
public static String doit(String input, Map<String, String> replacements)
{
String patternString = join(replacements.keySet(), "|");
Pattern pattern = Pattern.compile(patternString);
Matcher matcher = pattern.matcher(input);
StringBuffer resultStringBuffer = new StringBuffer();
while (matcher.find())
{
System.out.printf("match found: %s at start: %d, end: %d\n",
matcher.group(), matcher.start(), matcher.end());
String matchedPattern = matcher.group();
String replaceWith = replacements.get(matchedPattern);
// Do the replacement here.
matcher.appendReplacement(resultStringBuffer, replaceWith);
}
matcher.appendTail(resultStringBuffer);
return resultStringBuffer.toString();
}
private static String join(Set<String> set, String delimiter)
{
StringBuilder sb = new StringBuilder();
int numElements = set.size();
int i = 0;
for (String s : set)
{
sb.append(Pattern.quote(s));
if (i++ < numElements-1) { sb.append(delimiter); }
}
return sb.toString();
}
}
This prints out:
match found: i at start: 1, end: 2
match found: i at start: 4, end: 5
fIt I am
Ideally, it should be fIt EYE am.

You mistyped one of your regexes:
replacements.put("\\bi\\", "EYE"); //Should be \\bi\\b
replacements.put("i", "I");
You may also want to make your regexes unique. There is no guarantee of order with map.getKeySet() so it may just be replacing i with I before checking \\bi\\b.

You could use capture groups, without straying too far from your existing design. So instead of using the matched pattern as the key, you look up based on the order within a List.
You would need to change the join method to put parantheses around each of the patterns, something like this:
private static String join(Set<String> set, String delimiter) {
StringBuilder sb = new StringBuilder();
sb.append("(");
int numElements = set.size();
int i = 0;
for (String s : set) {
sb.append(s);
if (i++ < numElements - 1) {
sb.append(")");
sb.append(delimiter);
sb.append("("); }
}
sb.append(")");
return sb.toString();
}
As a side note, the use of Pattern.quote in the original code listing would have caused the match to fail where those metacharacters were present.
Having done this, you would now need to determine which of the capture groups was responsible for the match. For simplicity I'm going to assume that none of the match patterns will themselves contain capture groups, in which case something like this would work, within the matcher while loop:
int index = -1;
for (int j=1;j<=replacements.size();j++){
if (matcher.group(j) != null) {
index = j;
break;
}
}
if (index >= 0) {
System.out.printf("Match on index %d = %s %d %d\n", index, matcher.group(index), matcher.start(index), matcher.end(index));
}
Next, we would like to use the resulting index value to index straight back into the replacements. The original code uses a HashMap, which is not suitable for this; you're going to have to refactor that to use a pair of Lists in some form, one containing the list of match patterns and the other the corresponding list of replacement strings. I won't do that here, but I hope that provides enough detail to create a working solution.

Most efficient way to extract all the (natural) numbers from a string

Users may want to delimit numbers as they want.
What is the most efficient (or a simple standard function) to extract all the (natural) numbers from a string?

You could use a regular expression. I modified this example from Sun's regex matcher tutorial:
import java.util.regex.Pattern;
import java.util.regex.Matcher;
public class Test {
private static final String REGEX = "\\d+";
private static final String INPUT = "dog dog 1342 dog doggie 2321 dogg";
public static void main(String[] args) {
Pattern p = Pattern.compile(REGEX);
Matcher m = p.matcher(INPUT); // get a matcher object
while(m.find()) {
System.out.println("start(): "+m.start());
System.out.println("end(): "+m.end());
}
}
}
It finds the start and end indexes of each number. Numbers starting with 0 are allowed with the regular expression \d+, but you could easily change that if you want to.

I'm not sure I understand your question exactly. But if all you want is to pull out all non-negative integers then this should work pretty nicely:
String foo = "12,34,56.0567 junk 6745 some - stuff tab tab 789";
String[] nums = foo.split("\\D+");
// nums = ["12", "34", "56", "0567", "6745", "789"]
and then parse out the strings as ints (if needed).

If you know the delimiter, then:
String X = "12,34,56";
String[] y = X.split(","); // d=delimiter
int[] z = new int[y.length];
for (int i = 0; i < y.length; i++ )
{
z[i] = java.lang.Integer.valueOf(y[i]).intValue();
}
If you don't, you probably need to pre-process - you could do x.replace("[A-Za-z]", " "); and replace all characters with spaces and use space as the delimiter.
Hope that helps - I don't think there is a built-in function.

Is there a way to split strings with String.split() and include the delimiters? [duplicate]

I have a multiline string which is delimited by a set of different delimiters:
(Text1)(DelimiterA)(Text2)(DelimiterC)(Text3)(DelimiterB)(Text4)
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?

You can use lookahead and lookbehind, which are features of regular expressions.
System.out.println(Arrays.toString("a;b;c;d".split("(?<=;)")));
System.out.println(Arrays.toString("a;b;c;d".split("(?=;)")));
System.out.println(Arrays.toString("a;b;c;d".split("((?<=;)|(?=;))")));
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, ";"));
...
}

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);
}
System.out.println();
}
public static void main(String[] args) {
dump("1,234,567,890".split(","));
// "[1][234][567][890]"
dump("1,234,567,890".split("(?=,)"));
// "[1][,234][,567][,890]"
dump("1,234,567,890".split("(?<=,)"));
// "[1,][234,][567,][890]"
dump("1,234,567,890".split("(?<=,)|(?=,)"));
// "[1][,][234][,][567][,][890]"
dump(":a:bb::c:".split("(?=:)|(?<=:)"));
// "[][:][a][:][bb][:][:][c][:]"
dump(":a:bb::c:".split("(?=(?!^):)|(?<=:)"));
// "[:][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]"
dump("ArrayIndexOutOfBoundsException".split("(?<=[a-z])(?=[A-Z])"));
// "[Array][Index][Out][Of][Bounds][Exception]"
dump("1234567890".split("(?<=\\G.{4})"));
// "[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
Java split is eating my characters.
Can you use zero-width matching regex in String split?
How do I convert CamelCase into human-readable names in Java?
Backreferences in lookbehind
See also
regular-expressions.info/Lookarounds

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.

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()) {
splitted.add(text.substring(last_match,m.start()));
if (this.keep_delimiters) {
splitted.add(m.group());
}
last_match = m.end();
}
splitted.add(text.substring(last_match));
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>");
return;
}
Pattern pattern = null;
try {
pattern = Pattern.compile(argv[0]);
}
catch (PatternSyntaxException e) {
System.err.println(e);
return;
}
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);
}
}
}
/*
Example:
> 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

Pass the 3rd aurgument as "true". It will return delimiters as well.
StringTokenizer(String str, String delimiters, true);

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] + '"');
}
OUTPUT:
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.

I got here late, but returning to the original question, why not just use lookarounds?
Pattern p = Pattern.compile("(?<=\\w)(?=\\W)|(?<=\\W)(?=\\w)");
System.out.println(Arrays.toString(p.split("'ab','cd','eg'")));
System.out.println(Arrays.toString(p.split("boo:and:foo")));
output:
[', 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.

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:
https://stackoverflow.com/users/18393/cletus
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()));
ret.add(m.group());
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);
}
System.out.println();
}
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"
}
}

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,".split(",")
returns
"", "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!
Splitter.on(',')
.trimResults()
.omitEmptyStrings()
.split(" foo, ,bar, quux,")
--> ["foo", "bar", "quux"]
Order of config methods doesn't matter -- during splitting, trimming happens before checking for empties.

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());
result.add(matcher.group());
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:
#Test
public void splitsVariableLengthPattern() {
String[] result = Split.split("/foo/$bar/bas", "\\$\\w+");
Assert.assertArrayEquals(new String[] { "/foo/", "$bar", "/bas" }, result);
}
#Test
public void splitsEndingWithPattern() {
String[] result = Split.split("/foo/$bar", "\\$\\w+");
Assert.assertArrayEquals(new String[] { "/foo/", "$bar" }, result);
}
#Test
public void splitsStartingWithPattern() {
String[] result = Split.split("$foo/bar", "\\$\\w+");
Assert.assertArrayEquals(new String[] { "", "$foo", "/bar" }, result);
}
#Test
public void splitsNoMatchesPattern() {
String[] result = Split.split("/foo/bar", "\\$\\w+");
Assert.assertArrayEquals(new String[] { "/foo/bar" }, result);
}

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;
while(matcher.find()){
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);
list.add(finalElement);
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();
while(matcher.find()){
matcher.appendReplacement(stringBuffer, matcher.group());
list.add(stringBuffer.toString());
stringBuffer.setLength(0); //clear buffer
}
matcher.appendTail(stringBuffer); ///dodajemy reszte ciagu
list.add(stringBuffer.toString());
return list.toArray(new String[list.size()]);
}

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()) {
tokens.add(tokenMatcher.group());
} else {
// report error
break;
}
}
System.out.println(tokens);
}
}
Sample output:
[58.5, +, variable, -, +, 98, *, 78, /, 96, +, a, /, 78.7, -, 3443, *, 12, -, 3]

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) {
result.add(sb.toString());
}
sb = new StringBuilder();
wasLetterOrDigit = !wasLetterOrDigit;
}
sb.append(c);
}
result.add(sb.toString());
return result.toArray(new String[0]);
}

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.

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

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.

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 + "|");
System.out.println();
Result:
'|ab|','|cd|','|eg|'|
A bit too much... :-)

Tweaked Pattern.split() to include matched pattern to the list
Added
// 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();
matchList.add(match);
// 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())
.toString();
matchList.add(match);
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(""))
resultSize--;
String[] result = new String[resultSize];
return matchList.subList(0, resultSize).toArray(result);
}

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()
lastMatch=m.end()
}
if (lastMatch < str.length()) tokens << str[lastMatch..<str.length()]
tokens
}
[['<html><head><title>this is the title</title></head>',/<[^>]+>/],
['before<html><head><title>this is the title</title></head>after',/<[^>]+>/]
].each {
println splitWithTokens(*it)
}

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)
System.out.println(string);
String temp3[]=new String[temp.length-1];
for(String string:temp2)
{
System.out.println(string);
if((string.equals("")!=true)&&(string.equals("\\s")!=true))
{
temp3[i]=string;
i++;
}
// System.out.println(temp.length);
// System.out.println(temp2.length);
}
System.out.println(temp3.length);
String[] temp4=new String[temp.length+temp3.length];
int j=0;
for(i=0;i<temp.length;i++)
{
temp4[j]=temp[i];
j=j+2;
}
j=1;
for(i=0;i<temp3.length;i++)
{
temp4[j]=temp3[i];
j+=2;
}
for(String s:temp4)
System.out.println(s);

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("~");
System.out.println(Arrays.toString(expressionArray));

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" )
}

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(w+"t");
}
System.out.println("Split array length: "+arr.length);
System.out.println("method 2");
System.out.println(str.replaceAll("t", "\n"+"t"));
}

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.