Solution:

value.replaceAll("[^A-Za-z0-9]", "")

Explanation:

[^abc] When a caret ^ appears as the first character inside square brackets, it negates the pattern. This pattern matches any character except a or b or c.

Looking at the keyword as two function:

• [(Pattern)] = match(Pattern)
• [^(Pattern)] = notMatch(Pattern)

Moreover regarding a pattern:

• A-Z = all characters included from A to Z

• a-z = all characters included from a to z

• 0=9 = all characters included from 0 to 9

Therefore it will substitute all the char NOT included in the pattern

If you want to also allow alphanumeric characters which don't belong to the ascii characters set, like for instance german umlaut's, you can consider using the following solution:

 String value = "your value";

 // this could be placed as a static final constant, so the compiling is only done once
 Pattern pattern = Pattern.compile("[^\\w]", Pattern.UNICODE_CHARACTER_CLASS);

 value = pattern.matcher(value).replaceAll("");

Please note that the usage of the UNICODE_CHARACTER_CLASS flag could have an impose on performance penalty (see javadoc of this flag)

Using Guava you can easily combine different type of criteria. For your specific solution you can use:

value = CharMatcher.inRange('0', '9')
        .or(CharMatcher.inRange('a', 'z')
        .or(CharMatcher.inRange('A', 'Z'))).retainFrom(value)

Guava's CharMatcher provides a concise solution:

output = CharMatcher.javaLetterOrDigit().retainFrom(input);