That depends on what you mean. If you just want to get rid of them, do this:
(Update: Apparently you want to keep digits as well, use the second lines in that case)

String alphaOnly = input.replaceAll("[^a-zA-Z]+","");
String alphaAndDigits = input.replaceAll("[^a-zA-Z0-9]+","");

or the equivalent:

String alphaOnly = input.replaceAll("[^\\p{Alpha}]+","");
String alphaAndDigits = input.replaceAll("[^\\p{Alpha}\\p{Digit}]+","");

(All of these can be significantly improved by precompiling the regex pattern and storing it in a constant)

Or, with Guava:

private static final CharMatcher ALNUM =
  CharMatcher.inRange('a', 'z').or(CharMatcher.inRange('A', 'Z'))
  .or(CharMatcher.inRange('0', '9')).precomputed();
// ...
String alphaAndDigits = ALNUM.retainFrom(input);

But if you want to turn accented characters into something sensible that's still ascii, look at these questions:

• Converting Java String to ASCII
• Java change áéőűú to aeouu
• ń ǹ ň ñ ṅ ņ ṇ ṋ ṉ ̈ ɲ ƞ ᶇ ɳ ȵ --> n or Remove diacritical marks from unicode chars

This worked for me:

String result = str.replaceAll("[^\\dA-Za-z ]", "").replaceAll("\\s+", "+");

For this input string:

/-+!@#$%^&())";:[]{}\ |wetyk 678dfgh

It yielded this result:

+wetyk+678dfgh

use [\\W+] or "[^a-zA-Z0-9]" as regex to match any special characters and also use String.replaceAll(regex, String) to replace the spl charecter with an empty string. remember as the first arg of String.replaceAll is a regex you have to escape it with a backslash to treat em as a literal charcter.

          String c= "hjdg$h&jk8^i0ssh6";
        Pattern pt = Pattern.compile("[^a-zA-Z0-9]");
        Matcher match= pt.matcher(c);
        while(match.find())
        {
            String s= match.group();
        c=c.replaceAll("\\"+s, "");
        }
        System.out.println(c);

If I understood your problem correctly then this problem seems to be tricky. You can try this code to solve your task:

// your special characters
String regex = "+ – && || ! ( ) { } [ ] ^ ” ~ * ? : \\";
// building a valid regex out of above
regex = '(' + regex.replaceAll("([^\\s]{1,2})(?=(?:\\s+|1\\\\E").replace(' ', '|') + ')';

// your string to be replaced
String str = "content:you&&me";
// actual replacement
str = str.replaceAll(regex, "\\\\$1");

// printing the result
System.out.printf("********* replaced: [%s]%n", str);

Live Demo: http://ideone.com/onmcMy

Simply use String#replace(CharSequence target, CharSequence replacement) in your case to replace a given CharSequence, as next:

special = special.replace("@$", "as");

Or use Pattern.quote(String s) to convert your String as a literal pattern String, as next:

special = special.replaceAll(Pattern.quote("@$"), "as");

If you intend to do it very frequently, consider reusing the corresponding Pattern instance (the class Pattern is thread-safe which means that you can share instances of this class) to avoid compiling your regular expression at each call which has a price in term of performances.

So your code could be:

private static final Pattern PATTERN = Pattern.compile("@$", Pattern.LITERAL);
...
special = PATTERN.matcher(special).replaceAll("as");

You could use a lookahead, to find the end of the string, preceded by characters other than dot.

yourString.replaceAll("\\.(?=[^.]*$)", "replacement");

Note that the first dot needs to be escaped with a backslash, because dot has a special meaning in a regular expression (it matches any character). The second dot doesn't need to be escaped, because the special meaning doesn't apply in square brackets.

The (?= ) structure means "followed by this" - in other words, the dot that you match can be followed by any number of non-dot characters, and then the end of the string. Those extra characters are not considered part of the match.

Don't use square brackets, as it represents a set of single characters to match (a character class).

a=a.replaceAll("’|‵", "-");

Demo!

You need to:

• replace all commas within quotes with +
• replace non-whitelist (and you need to add commas to your whitelist) +
• remove double quotes

Try this:

line = line.replaceAll("[^A-Za-z0-9\",]|,(?!(([^\"]*\"){2})*[^\"]*$)", "+").replace("\"", "");

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're close, you just need to add a quantifier to the expression to allow it to match more than one character.

/[^A-Za-z0-9]+/

(Note the + at the end.)

So, your code should be this:

"Town & Country".replaceAll("[^A-Za-z0-9]+", "-").toLowerCase();

Live example on regexr

You can get rid of all characters outside the printable ASCII range using String#replaceAll() by replacing the pattern [^\\x20-\\x7e] with an empty string:

a = a.replaceAll("[^\\x20-\\x7e]", "");

But this actually doesn't solve your actual problem. It's more a workaround. With the given information it's hard to nail down the root cause of this problem, but reading either of those articles must help a lot:

• The Absolute Minimum Every Software Developer Absolutely, Positively Must Know About Unicode and Character Sets (No Excuses!)
• Unicode - How to get the characters right?

Use replaceAll("[^\\w\\s\\-_]", "");

What I did was add the underscore and hyphen to the regular expression. I added a \\ before the hyphen because it also serves for specifying ranges: a-z means all letters between a and z. Escaping it with \\ makes sure it is treated as an hyphen.