替换不在范围内的所有字符（Java String）

Question

How do you replace all of the characters in a string that do not fit a criteria. I'm having trouble specifically with the NOT operator.

Specifically, I'm trying to remove all characters that are not a digit, I've tried this so far:

String number = "703-463-9281";
String number2 = number.replaceAll("[0-9]!", ""); // produces: "703-463-9281" (no change)
String number3 = number.replaceAll("[0-9]", "");  // produces: "--" 
String number4 = number.replaceAll("![0-9]", ""); // produces: "703-463-9281" (no change)
String number6 = number.replaceAll("^[0-9]", ""); // produces: "03-463-9281"

Answer 1

To explain: The ^ at the start of a character class will negate that class But it has to be inside the class for that to work. The same character outside a character class is the anchor for start of string/line instead.

You can try this instead:

"[^0-9]"

Answer 2

Here's a quick cheat sheet of character class definition and how it interacts with some regex meta characters.

• [aeiou] - matches exactly one lowercase vowel
• [^aeiou] - matches a character that ISN'T a lowercase vowel ( negated character class)
• ^[aeiou] - matches a lowercase vowel anchored at the beginning of the line
• [^^] - matches a character that isn't a caret/ '^'
• ^[^^] - matches a character that isn't a caret at the beginning of line
• ^[^.]. - matches anything but a literal period, followed by "any" character, at the beginning of line
• [az] - matches exactly one character within the range of 'a' to 'z' (ie all lowercase letters)
• [az-] - matches either an 'a' , a 'z' , or a '-' (literal dash)
• [.*]* - matches a contiguous sequence (possibly empty) of dots and asterisks
• [aeiou]{3} - matches 3 consecutive lowercase vowels (all not necessarily the same vowel)
• \\[aeiou\\] - matches the string "[aeiou]"

References

• regular-expressions.info/Character class , Anchors , Dot , Repetition

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