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Splitting string in java produces empty element first

Im trying to split a sting on multiple or single occurences of "O" and all other characters will be dots. I'm wondering why this produces en empty string first.
String row = ".....O.O.O"
String[] arr = row.split("\\.+");
This produces produces:
["", "O", "O", "O"]

You just need to make sure that any trailing or leading dots are removed.
So one solution is:
row.replaceAll("^\\.+|\\.+$", "").split("\\.+");

For this pattern you can use replaceFirstMethod() and then split by dot
String[] arr = row.replaceFirst("\\.+","").split("\\.");
Output will be
["O","O","O"]

The "+" character is removing multiple instances of the seperator, so what your split is essentially doing is splitting the following string on "."
.0.0.0.
This, of course, means that your first field is empty. Hence the result you get.
To avoid this, strip all leading separators from the string before splitting it. Rather than type some examples on how to do this, here's a thread with a few suggestions.
Java - Trim leading or trailing characters from a string?

Java - What is the right regex for my String.split()

I am splitting equation string into string array like this:
String[] equation_array = (equation.split("(?<=[-+×÷)(])|(?=[-+×÷)(])"));
Now for test string:
test = "4+(2×5)"
result is fine:
test_array = {"4", "+", "(", "2",...}
but for test string:
test2 = "(2×5)+5"
I got string array:
test2_array = {"", "(", "×",...}.
So, problem is why does it add an empty string before ( in array after splitting?

This is actually known behavior in Java regex.
To avoid this empty result use this negative lookahead based regex:
String[] equation_array = "(2×5)+5".split("(?!^)((?<=[-+×÷)(])|(?=[-+×÷)(]))");
//=> ["(", "2", "×", "5", ")", "+", "5"]
What (?!^) means is to avoid splitting at line start.

You can add condition that not to split if before token is start of string like
"(?<=[-+×÷)(])|(?<!^)(?=[-+×÷)(])"
^^^^^^

What about looking backwards to make sure we're not at the start of the string, and looking forwards to make sure we're not at the end?
"(?<=[-+×÷)(])(?!$)|(?<!^)(?=[-+×÷)(])"
Here ^ and $ are start and end of string indicators and (?!...) and (?<!...) are negative lookahead and lookbehind.

problem is why does it add an empty string before ( in array after splitting?
Because for the input (2×5)+5 the regex used for splitting matches right at the start-of-string because of the positive look ahead (?=[-+×÷)(]).
(2×5)+5
↖
It matches right here before the (, resulting in an empty string: "".
My advice would be not to use regular expressions to parse mathematical expressions, there are more suitable algorithms for this.

Java (Regex) - Get all words in a sentence

I need to split a java string into an array of words. Let's say the string is:
"Hi!! I need to split this string, into a serie's of words?!"
At the moment I'm tried using this String[] strs = str.split("(?!\\w)") however it keeps symbols such as ! in the array and it also keeps strings like "Hi!" in the array as well. The string I am splitting will always be lowercase. What I would like is for an array to be produced that looks like:
{"hi", "i", "need", "to", "split", "this", "string", "into", "a", "serie's", "of", "words"} - Note the apostrophe is kept.
How could I change my regex to not include the symbols in the array?
Apologies, I would define a word as a sequence of alphanumeric characters only but with the ' character inclusive if it is in the above context such as "it's", not if it is used to a quote a word such as "'its'". Also, in this context "hi," or "hi-person" are not words but "hi" and "person" are. I hope that clarifies the question.

You can remove all ?! symbols and then split into words
str = str.replaceAll("[!?,]", "");
String[] words = str.split("\\s+");
Result:
Hi, I, need, to, split, this, string, into, a, serie's, of, words

Should work for what you want.
String line = "Hi!! I need to split this string, into a serie's of words?! but not '' or ''' word";
String regex = "([^a-zA-Z']+)'*\\1*";
String[] split = line.split(regex);
System.out.println(Arrays.asList(split));
Gives
[Hi, I, need, to, split, this, string, into, a, serie's, of, words, but, not, or, word]

If you define a word as a sequence of non-whitespace characters (whitespace character as defined by \s), then you can split along space characters:
str.split("\\s+")
Note that ";.';.##$>?>#4", "very,bad,punctuation", and "'goodbye'" are words under the definition above.
Then the other approach is to define a word as a sequence of characters from a set of allowed characters. If you want to allow a-z, A-Z, and ' as part of a word, you can split along everything else:
str.split("[^a-zA-Z']+")
This will still allow "''''''" to be defined as a word, though.

So what you want is to split on anything that is not a wordcharacter [a-zA-Z] and is not a '
This regex will do that "[^a-zA-Z']\s"
There will be a problem if the string contains a quote that is quoted in '
I usually use this page for testing my regex'
http://www.regexplanet.com/advanced/java/index.html

I would use str.split("[\\s,?!]+"). You can add whatever character you want to split with inside the brackets [].

You could filter out the characters you deem as "non-word" characters:
String[] strs = str.split("[,!? ]+");

myString.replaceAll("[^a-zA-Z'\\s]","").toLowerCase().split("\\s+");
replaceAll("[^a-zA-Z'\\s]","") method replaces all the characters which are not a-z or A-Z or ' or a whitespace with nothing ("") and then toLowerCase method make all the chars returned from replaceAll method lower case. Finally we are splitting the string in terms of whitespace char. more readable one;
myString = myString.replaceAll("[^a-zA-Z'\\s]","");
myString = myString.toLowerCase();
String[] strArr = myString.split("\\s+");

Matching strings and storing into an array with regex in java

Im making a program that takes a file and finds identifiers. So far I removed any words in quotes, any words that start with a number and I removed all the non word characters.
Is there a way to find words that dont match words in an array and store those words into another array using regex? I can figure it out, I was trying to use the split method but its not working right when I try to split by spaces...This is what I did to split it.
String[] SplitString = newLine.split("[\\s]");

Use
String[] SplitString = newLine.split("\\s");
if you don't want to combine multiple spaces/tabs, etc., but use
String[] SplitString = newLine.split("\\s+");
if you do. For example, if your string is:
"a b c"
the first will give you four tokens: "a", "", "b", and "c", and the second will give you three: "a", "b", and "c".

You can do it simply in one line by first removing the known words, then splitting:
String[] unknownWords = newLine.replaceAll("\\b(apple|orange|banana)\\b", "").split("\\s+");
Notes:
Your regex [\s] is equivalent to \s, so I simplified it
You should probably split on any number of spaces: \s+
\b means "word boundary" - this means the removal regex won't match applejack
The regex (A|B|C|etc) is the syntax for "OR" logic

How to split a String in Java, ignoring multiple successive tokens

I'm trying to parse arguments for a command, but if I were to put multiple spaces in a row, String.split() will leave empty Strings in the result array. Is there a way I can get rid of this?
For example: "abc 123".split(" ") results in {"abc", "", "", "", "", "123"} but what I really want is {"abc", "123"}

Just use regex
"abc 123".split("\\s+");
Here \s is any whitespace character and \s+ is one or more consecutive whitespace characters.