替换所有特殊字符，数字和字母

Question

This is the code for example

String a="abcd ABCD 0123 !@#$%^&*()";

This is syntax for replacing numbers & alphabets with blank

a = a.replaceAll("[a-zA-Z0-9]","");

This is syntax for replacing special characters with blank

a = a.replaceAll("[^\\w\\s-_]","");

So how do I combine both of these syntax to replace special characters, numbers and alphabets with blank without using this method a = a.replaceAll(".",""); to replace entire string with blank?

Is there any other way ??

Answer 1

I am not sure what you want to achieve, but if you want to combine two character sets you can

• use OR operator like [oneSet]|[secondSet]
• place them in other set like [[oneSet][secondSet]]
  • which can also be simplified to [oneSet[secondSet]] so maybe you are looking for something like

a = a.replaceAll("[a-zA-Z0-9[^\\w\\s-_]]","");

---

Demo: I will replace found characters with X to show exactly which characters ware replaced

String a = "abcd ABCD 0123 !@#$%^&*()";
System.out.println(a.replaceAll("[a-zA-Z0-9]", "X"));
System.out.println(a.replaceAll("[^\\w\\s-_]", "X"));
System.out.println(a.replaceAll("[a-zA-Z0-9[^\\w\\s-_]]", "X"));

Output:

XXXX XXXX XXXX !@#$%^&*()
abcd ABCD 0123 XXXXXXXXXX
XXXX XXXX XXXX XXXXXXXXXX

Answer 2

Combine with | :

a=a.replaceAll("[a-zA-Z0-9]|[^\\w\\s-_]", "");

Answer 3

Alternatively you can use below to account for line-breaks

String s = "abcd ABCD 0123 !@#$%^&*()";
System.out.println(s.replaceAll("(?s)."," "));