How to split a string with any whitespace chars as delimiters

Question

What regex pattern would need I to pass to java.lang.String.split() to split a String into an Array of substrings using all whitespace characters (' ', '\t', '\n', etc.) as delimiters?

Answer 1

Something in the lines of

myString.split("\\s+"); 

This groups all white spaces as a delimiter.

So if I have the string:

"Hello[space character][tab character]World" 

This should yield the strings "Hello" and "World" and omit the empty space between the [space] and the [tab].

As VonC pointed out, the backslash should be escaped, because Java would first try to escape the string to a special character, and send that to be parsed. What you want, is the literal "\s", which means, you need to pass "\\s". It can get a bit confusing.

The \\s is equivalent to [ \\t\\n\\x0B\\f\\r].

Answer 2

In most regex dialects there are a set of convenient character summaries you can use for this kind of thing - these are good ones to remember:

\w - Matches any word character.

\W - Matches any nonword character.

\s - Matches any white-space character.

\S - Matches anything but white-space characters.

\d - Matches any digit.

\D - Matches anything except digits.

A search for "Regex Cheatsheets" should reward you with a whole lot of useful summaries.