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Split String into words by multiple arguments for example both whitespace and "-"

So I want to split a string using by space and "-".
For example, a sample string : "White-spaces are necessary"
The above String should be split into parts like "White", "space", "are", "necessary"
is this possible using split() method in Java.

The split method takes a String in the form of a regex.
Using regex character classes, we can say split on either space OR dashes.
the regex would be the following [ -].
Each character inside the brackets is a possible character to split on.
However, this would split groups of 2 or more white space characters as separate split items, so you would end up with empty array elements. To avoid this you can use:
[ -]+
More reading on regex character classes.
https://www.regular-expressions.info/charclass.html

Java regex - split string with leading special characters

I am trying to split a string that contains whitespaces and special characters. The string starts with special characters.
When I run the code, the first array element is an empty string.
String s = ",hm ..To?day,.. is not T,uesday.";
String[] sArr = s.split("[^a-zA-Z]+\\s*");
Expected result is ["hm", "To", "day", "is", "not", "T", "uesday"]
Can someone explain how this is happening?
Actual result is ["", "hm", "To", "day", "is", "not", "T", "uesday"]

Split is behaving as expected by splitting off a zero-length string at the start before the first comma.
To fix, first remove all splitting chars from the start:
String[] sArr = s.replaceAll("^([^a-zA-Z]*\\s*)*", "").split("[^a-zA-Z]+\\s*");
Note that I’ve altered the removal regex to trim any sequence of spaces and non-letters from the front.
You don’t need to remove from the tail because split discards empty trailing elements from the result.

I would simplify it by making it a two-step process rather than trying to achieve a pure regex split() operation:
s.replaceAll( '[^a-zA-Z]+', ' ' ).trim().split( ' ' )

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+");

Split string based on regex but keep delimiters

I'm trying to split a string using a variety of characters as delimiters and also keep those delimiters in their own array index. For example say I want to split the string:
if (x>1) return x * fact(x-1);
using '(', '>', ')', '*', '-', ';' and '\s' as delimiters. I want the output to be the following string array: {"if", "(", "x", ">", "1", ")", "return", "x", "*", "fact", "(", "x", "-", "1", ")", ";"}
The regex I'm using so far is
split("(?=(\\w+(?=[\\s\\+\\-\\*/<(<=)>(>=)(==)(!=)=;,\\.\"\\(\\)\\[\\]\\{\\}])))")
which splits at each word character regardless of whether it is followed by one of the delimiters. For example
test + 1
outputs {"t","e","s","t+","1"} instead of {"test+", "1"}
Why does it split at each character even if that character is not followed by one of my delimiters? Also is a regex which does this even possible in Java?
Thank you

Well, you can use lookaround to split at points between characters without consuming the delimiters:
(?<=[()>*-;\s])|(?=[()>*-;\s])
This will create a split point before and after each delimiter character. You might need to remove superfluous whitespace elements from the resulting array, though.
Quick PowerShell test (| marks the split points):
PS Home:\> 'if (x>1) return x * fact(x-1);' -split '(?<=[()>*-;\s])|(?=[()>*-;\s])' -join '|'
if| |(|x|>|1|)| |return| |x| |*| |fact|(|x|-|1|)|;|

How about this pattern?
(\w+)|([\p{P}\p{S}])

To answer your question, "Why?", it's because your entire expression is a lookahead assertion. As long as that assertion is true at each character (or maybe I should say "between"), it is able to split.
Also, you cannot group within character classes, e.g. (<=) is not doing what you think it is doing.

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