Replace any sequence of non-letters with a single white space:

str.replaceAll("[^a-zA-Z]+", " ")

You also might want to apply trim() after the replace.

If you want to support languages other than English, use "[^\\p{IsAlphabetic}]+" or "[^\\p{IsLetter}]+". See this question about the differences.

You can remove everything but the digits:

phone = phone.replaceAll("[^0-9]","");

That depends on what you define as special characters, but try replaceAll(...):

String result = yourString.replaceAll("[-+.^:,]","");

Note that the ^ character must not be the first one in the list, since you'd then either have to escape it or it would mean "any but these characters".

Another note: the - character needs to be the first or last one on the list, otherwise you'd have to escape it or it would define a range ( e.g. :-, would mean "all characters in the range : to ,).

So, in order to keep consistency and not depend on character positioning, you might want to escape all those characters that have a special meaning in regular expressions (the following list is not complete, so be aware of other characters like (, {, $ etc.):

String result = yourString.replaceAll("[\\-\\+\\.\\^:,]","");

If you want to get rid of all punctuation and symbols, try this regex: \p{P}\p{S} (keep in mind that in Java strings you'd have to escape back slashes: "\\p{P}\\p{S}").

A third way could be something like this, if you can exactly define what should be left in your string:

String  result = yourString.replaceAll("[^\\w\\s]","");

This means: replace everything that is not a word character (a-z in any case, 0-9 or _) or whitespace.

Edit: please note that there are a couple of other patterns that might prove helpful. However, I can't explain them all, so have a look at the reference section of regular-expressions.info.

Here's less restrictive alternative to the "define allowed characters" approach, as suggested by Ray:

String  result = yourString.replaceAll("[^\\p{L}\\p{Z}]","");

The regex matches everything that is not a letter in any language and not a separator (whitespace, linebreak etc.). Note that you can't use [\P{L}\P{Z}] (upper case P means not having that property), since that would mean "everything that is not a letter or not whitespace", which almost matches everything, since letters are not whitespace and vice versa.

Additional information on Unicode

Some unicode characters seem to cause problems due to different possible ways to encode them (as a single code point or a combination of code points). Please refer to regular-expressions.info for more information.

st.replaceAll("\\s+","") removes all whitespaces and non-visible characters (e.g., tab, \n).

st.replaceAll("\\s+","") and st.replaceAll("\\s","") produce the same result.

The second regex is 20% faster than the first one, but as the number consecutive spaces increases, the first one performs better than the second one.

Assign the value to a variable, if not used directly:

st = st.replaceAll("\\s+","")