如何计算数字或字符匹配的模式？

Question

Here is my function that checks whether the string is alpha/numeric

public boolean bpIsAlphaNumeric(String s){
    String pattern= "^[a-zA-Z0-9]*$";
    if(s.matches(pattern)){
        return true;
    }
    return false; 
}

It works perfect. Now I need (using same method/pattern) to check that String s contains at least xxx alpha/numeric characters.

//e.g. at least 4 characters should be alphanumeric

String "abc#DE$01%23!##^$"  //true, because it contains - abcDE0123
String "!$#%#a#b&9$^$##^$"  //false, because it contains only - ab9

Answer 1

This is mostly from @f1sh's answer:

String regex = "([^a-zA-Z0-9]*[a-zA-Z0-9]){4,}"

It accepts ((non-alpha)* (alpha) at least 4 times).

Answer 2

I am not sure if the above pattern works. In case it does not, you can try this one as well:

public static boolean hasAtleast4AlphaNumeric(String s){
    String pattern= "([a-zA-Z0-9].*){4,}.*$";
    if(s.matches(pattern)){
        return true;
    }
    return false; 
}

System.out.println("Has four alphanumeric characters: " + hasAtleast4AlphaNumeric("abc#DE$01%23!##^$"));
System.out.println("Does not have four alphanumeric characters: " + hasAtleast4AlphaNumeric("!$#%#a#b&9$^$##^$"));
System.out.println("Does not have four alphanumeric characters: " + hasAtleast4AlphaNumeric("VVVV!$#%#a#b&9$^$##^$"));


Output:
    Has four alphanumeric characters: true
    Does not have four alphanumeric characters: false
    Has four alphanumeric characters: true

Use curly braces ({}) when you want to be very specific about the number of occurrences an operator or subexpression must match in the source string. Curly braces and their contents are known as interval expressions. You can specify an exact number or a range.