检查Java正则表达式中的特定字符

Question

How check if a String contains only one specific character? Eg:

On the String square/retrofit and square/retrofit/issues I need to check if the String has more than one / character.

square/retrofit/issues need to be false because have more than one / character and square/retrofit need to be true.

The string can have numbers.

Answer 1

You do not need regex. Simple indexOf and lastIndexOf methods should be enough.

boolean onlyOne = s.indexOf('/') == s.lastIndexOf('/');

EDIT 1
Of course, if / does not appear in given string above will be true . So, to avoid this situation you can also check what is returned index from one of these methods.

EDIT 2
Working solution:

class Strings {
    public static boolean availableOnlyOnce(String source, char c) {
        if (source == null || source.isEmpty()) {
            return false;
        }

        int indexOf = source.indexOf(c);
        return (indexOf == source.lastIndexOf(c)) && indexOf != -1;
    }
}

Test cases:

System.out.println(Strings.availableOnlyOnce("path", '/'));
System.out.println(Strings.availableOnlyOnce("path/path1", '/'));
System.out.println(Strings.availableOnlyOnce("path/path1/path2", '/'));

Prints:

false
true
false

Answer 2

Or if you'd like to use a bit more modern approach with streams:

boolean occursOnlyOnce(String stringToCheck, char charToMatch) {
    return stringToCheck.chars().filter(ch -> ch == charToMatch).count() == 1;
}

Disclaimer: This is not supposed to be the most optimal approach.

A bit more optimized approach:

    boolean occursOnlyOnce(String stringToCheck, char charToMatch) {
    boolean isFound = false;
    for (char ch : stringToCheck.toCharArray()) {
        if (ch == charToMatch) {
            if (!isFound) {
                isFound = true;
            } else {
                return false; // More than once, return immediately
            }
        }
    }
    return isFound; // Not found
}