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Weird behavior of Java's String.split() [duplicate]

I am trying to split the Value using a separator.
But I am finding the surprising results
String data = "5|6|7||8|9||";
String[] split = data.split("\\|");
System.out.println(split.length);
I am expecting to get 8 values. [5,6,7,EMPTY,8,9,EMPTY,EMPTY]
But I am getting only 6 values.
Any idea and how to fix. No matter EMPTY value comes at anyplace, it should be in array.

split(delimiter) by default removes trailing empty strings from result array. To turn this mechanism off we need to use overloaded version of split(delimiter, limit) with limit set to negative value like
String[] split = data.split("\\|", -1);
Little more details:
split(regex) internally returns result of split(regex, 0) and in documentation of this method you can find (emphasis mine)
The limit parameter controls the number of times the pattern is applied and therefore affects the length of the resulting array.
If the limit n is greater than zero then the pattern will be applied at most n - 1 times, the array's length will be no greater than n, and the array's last entry will contain all input beyond the last matched delimiter.
If n is non-positive then the pattern will be applied as many times as possible and the array can have any length.
If n is zero then the pattern will be applied as many times as possible, the array can have any length, and trailing empty strings will be discarded.
Exception:
It is worth mentioning that removing trailing empty string makes sense only if such empty strings were created by the split mechanism. So for "".split(anything) since we can't split "" farther we will get as result [""] array.
It happens because split didn't happen here, so "" despite being empty and trailing represents original string, not empty string which was created by splitting process.

From the documentation of String.split(String regex):
This method works as if by invoking the two-argument split method with the given expression and a limit argument of zero. Trailing empty strings are therefore not included in the resulting array.
So you will have to use the two argument version String.split(String regex, int limit) with a negative value:
String[] split = data.split("\\|",-1);
Doc:
If the limit n is greater than zero then the pattern will be applied at most n - 1 times, the array's length will be no greater than n, and the array's last entry will contain all input beyond the last matched delimiter. If n is non-positive then the pattern will be applied as many times as possible and the array can have any length. If n is zero then the pattern will be applied as many times as possible, the array can have any length, and trailing empty strings will be discarded.
This will not leave out any empty elements, including the trailing ones.

String[] split = data.split("\\|",-1);
This is not the actual requirement in all the time. The Drawback of above is show below:
Scenerio 1:
When all data are present:
String data = "5|6|7||8|9|10|";
String[] split = data.split("\\|");
String[] splt = data.split("\\|",-1);
System.out.println(split.length); //output: 7
System.out.println(splt.length); //output: 8
When data is missing:
Scenerio 2: Data Missing
String data = "5|6|7||8|||";
String[] split = data.split("\\|");
String[] splt = data.split("\\|",-1);
System.out.println(split.length); //output: 5
System.out.println(splt.length); //output: 8
Real requirement is length should be 7 although there is data missing. Because there are cases such as when I need to insert in database or something else. We can achieve this by using below approach.
String data = "5|6|7||8|||";
String[] split = data.split("\\|");
String[] splt = data.replaceAll("\\|$","").split("\\|",-1);
System.out.println(split.length); //output: 5
System.out.println(splt.length); //output:7
What I've done here is, I'm removing "|" pipe at the end and then splitting the String. If you have "," as a seperator then you need to add ",$" inside replaceAll.

From String.split() API Doc:
Splits this string around matches of the given regular expression.
This method works as if by invoking the two-argument split method with
the given expression and a limit argument of zero. Trailing empty
strings are therefore not included in the resulting array.
Overloaded String.split(regex, int) is more appropriate for your case.

you may have multiple separators, including whitespace characters, commas, semicolons, etc. take those in repeatable group with []+, like:
String[] tokens = "a , b, ,c; ;d, ".split( "[,; \t\n\r]+" );
you'll have 4 tokens -- a, b, c, d
leading separators in the source string need to be removed before applying this split.
as answer to question asked:
String data = "5|6|7||8|9||";
String[] split = data.split("[\\| \t\n\r]+");
whitespaces added just in case if you'll have those as separators along with |

Java String split operation

When I split a string in java as shown in code below, it gives length as 3.
String data = "DK4..||VSDK4|";
String[] tokens = data.split("\\|");
System.out.println(tokens.length);
However it should be 4, what could be the reason behind this?

From the split method documentation:
Trailing empty strings are therefore not included in the resulting array.
This is because split with a single argument calls split with second argument as 0, which, according to that method's documentation
If n is zero then the pattern will be applied as many times as possible, the array can have any length, and trailing empty strings will be discarded.

See the doc of split(String regex):
Splits this string around matches of the given regular expression.
This method works as if by invoking the two-argument split method with
the given expression and a limit argument of zero. Trailing empty
strings are therefore not included in the resulting array.
You can use split(String regex, int limit) with a negative limit to get all:
String[] tokens = data.split("\\|", -1);
If n is non-positive then the pattern will be applied as many times as
possible and the array can have any length. If n is zero then the
pattern will be applied as many times as possible, the array can have
any length, and trailing empty strings will be discarded.

Confused with using split with multiple delimiters

I'm practicing reading input and then tokenizing it.
For example, if I have [882,337] I want to just get the numbers 882 and 337. I tried using the following code:
String test = "[882,337]";
String[] tokens = test.split("\\[|\\]|,");
System.out.println(tokens[0]);
System.out.println(tokens[1]);
System.out.println(tokens[2]);
It kind of works, the output is:
(blank line)
882
337
What I don't understand is why token[0] is empty? I would expect there to only be two tokens where token[0] = 882 and token[1] = 337.
I checked out some links but didn't find the answer.
Thanks for the help!

Split splits the given String. If you split "[882,337]" on "[" or "," or "]" then you actually have:
nothing
882
337
nothing
But, as you have called String.split(delimiter), this calls String.split(delimiter, limit) with a limit of zero.
From the documentation:
The limit parameter controls the number of times the pattern is applied and therefore affects the length of the resulting array. If the limit n is greater than zero then the pattern will be applied at most n - 1 times, the array's length will be no greater than n, and the array's last entry will contain all input beyond the last matched delimiter. If n is non-positive then the pattern will be applied as many times as possible and the array can have any length. If n is zero then the pattern will be applied as many times as possible, the array can have any length, and trailing empty strings will be discarded.
(emphasis mine)
So in this configuration the final, empty, strings are discarded. You are therefore left with exactly what you have.
Usually, to tokenize something like this, one would go for a combination of replaceAll and split:
final String[] tokens = input.replaceAll("^\\[|\\]$").split(",");
This will first strip off the start (^[) and end (]$) brackets and then split on ,. This way you don't have to have somewhat obtuse program logic where you start looping from an arbitrary index.
As an alternative, for more complex tokenizations, one can use Pattern - might be overkill here, but worth bearing in mind before you get into writing multiple replaceAll chains.
First we need to define, in Regex, the tokens we want (rather than those we're splitting on) - in this case it's simple, it's just digits so \d.
So, in order to extract all digit only (no thousands/decimal separators) values from an arbitrary String on would do the following:
final List<Integer> tokens = new ArrayList<>(); <-- to hold the tokens
final Pattern pattern = Pattern.compile("\\d++"); <-- the compiled regex
final Matcher matcher = pattern.matcher(input); <-- the matcher on input
while(matcher.find()) { <-- for each matched token
tokens.add(Integer.parseInt(matcher.group())); <-- parse and `int` and store
}
N.B: I have used a possessive regex pattern for efficiency
So, you see, the above code is somewhat more complex than the simple replaceAll().split(), but it is much more extensible. You can use arbitrary complex regex to token almost any input.

The symbols where the string is split are here:
String test = "[882,337]";
^ ^ ^
Because The first char matches your delimiter, everything left from it will be the first result. Well, left from the first letter is nothing, so the result is the empty string.
One could expect the same behaviour for the end, since the last symbol also matches the delimiter. But:
Trailing empty strings are therefore not included in the resulting array.
See Javadoc.

Splitting creates two (or more) things from one thing. For instance if you split a,b by , you will get a and b.
But in case of ",b" you will get "" and "b". You can think of it this way:
"" exists at start, end and even in-between all characters of string:
""+","+"b" -> ",b" so if we split on this "," we are getting left and right part: "" and "b"
Similar things happens in case of "a," and at first result array is ["a",""] but here split method removes trailing empty strings and returns only ["a"] (you can turn off this clearing mechanism by using split(",", -1)).
So in case of
String test = "[882,337]";
String[] tokens = test.split("\\[|\\]|,");
you are splitting:
""+"["+"882"+","+"337"+"]"+""
here: ^ ^ ^
which at first creates array ["", "882", "337", ""] but then trailing empty string is removed and finally you are receiving:
["", "882", "337"]
Only case where empty string is removed from start of result array is when
you are using Java 8 (or newer) and splitting on regex which is zero-length like split("") or lets say before each x with split("(?=x)") (more info at: Why in Java 8 split sometimes removes empty strings at start of result array?)
and when this empty string was result of split method. For instance "".split("") will not remove "", more info here: https://stackoverflow.com/a/25058091/1393766

That's because each delimiter has a "before" and "after" result, even if it is empty. Consider
882,337
You expect that to produce two results.
Similarly, you expect
882,337,
to produce three, with the last one being empty (assuming your limit is big enough, or assuming you're using almost any other language / implementation of split()). Extending that logically,
,882,337,
must produce four, with the first and last results being empty. This is exactly the case you have, except you have multiple delimiters.

Tokenizing a string using negations

So i have the following problem:
I have to tokenize a string using String.split() and the tokens must be in the form 07dd ddd ddd, where d is a digit. I thought of using the following regex : ^(07\\d{2}\\s\\d{3}\\d{3}) and pass it as an argument to String.split(). But for some reason, although i do have substrings under that form, it outputs the whole initial string and doesn't tokenize it.
I initially thought that it was using an empty string as a splitter, as an empty string indeed matches that regex, but even after I added & (.)+ to the regex in order to assure that the splitter hasn't got length 0, it still outputs the whole initial string.
I know that i could have used Pattern's and Matchers to solve it much faster, but i have to use String.split(). Any ideas why this happens?

A Few Pointers
Your pattern ^(07\d{2}\s\d{3}\d{3}) is missing a space between the two last groups of digits
The reason you get the whole string back is that this pattern was never found in the first place: there is no split
If you split on this pattern (once fixed), the resulting array will be strings that are in-between this pattern (these tokens are actually removed)
If you want to use this pattern (once fixed), you need a Match All not a Split. This will look like arrayOfMatches = yourString.match(/pattern/g);
If you want to split, you need to use a delimiter that is present between the token (this delimiter could in fact just be a zero-width position asserted by the 07 about to follow)
Further Reading
Match All and Split are Two Sides of the Same Coin

String.split() - matching leading empty String prior to first delimiter?

I need to be able to split an input String by commas, semi-colons or white-space (or a mix of the three). I would also like to treat multiple consecutive delimiters in the input as a single delimiter. Here's what I have so far:
String regex = "[,;\\s]+";
return input.split(regex);
This works, except for when the input string starts with one of the delimiter characters, in which case the first element of the result array is an empty String. I do not want my result to have empty Strings, so that something like, ",,,,ZERO; , ;;ONE ,TWO;," returns just a three element array containing the capitalized Strings.
Is there a better way to do this than stripping out any leading characters that match my reg-ex prior to invoking String.split?
Thanks in advance!

No, there isn't. You can only ignore trailing delimiters by providing 0 as a second parameter to String's split() method:
return input.split(regex, 0);
but for leading delimiters, you'll have to strip them first:
return input.replaceFirst("^"+regex, "").split(regex, 0);

If by "better" you mean higher performance then you might want to try creating a regular expression that matches what you want to match and using Matcher.find in a loop and pulling out the matches as you find them. This saves modifying the string first. But measure it for yourself to see which is faster for your data.
If by "better" you mean simpler, then no I don't think there is a simpler way than the way you suggested: removing the leading separators before applying the split.

Pretty much all splitting facilities built into the JDK are broken one way or another. You'd be better off using a third-party class such as Splitter, which is both flexible and correct in how it handles empty tokens and whitespaces:
Splitter.on(CharMatcher.anyOf(";,").or(CharMatcher.WHITESPACE))
.omitEmptyStrings()
.split(",,,ZERO;,ONE TWO");
will yield an Iterable<String> containing "ZERO", "ONE", "TWO"

You could also potentially use StringTokenizer to build the list, depending what you need to do with it:
StringTokenizer st = new StringTokenizer(",,,ZERO;,ONE TWO", ",; ", false);
while(st.hasMoreTokens()) {
String str = st.nextToken();
//add to list, process, etc...
}
As a caveat, however, you'll need to define each potential whitespace character separately in the second argument to the constructor.