How to Count Digits in a String in Java

The Challenge

Count how many numeric digits (0-9) are in a string!

Solution

Method 1: Using Character.isDigit()

java

String text = "Hello123World456";
int digitCount = 0;

for (int i = 0; i < text.length(); i++) {
    if (Character.isDigit(text.charAt(i))) {
        digitCount++;
    }
}

System.out.println("Digit count: " + digitCount);
// Output: Digit count: 6

Method 2: Using Regex

java

String text = "Hello123World456";
int digitCount = text.replaceAll("[^0-9]", "").length();

System.out.println("Digit count: " + digitCount);
// Output: Digit count: 6

Method 3: Using Java 8 Streams

java

String text = "Hello123World456";
long digitCount = text.chars()
                      .filter(Character::isDigit)
                      .count();

System.out.println("Digit count: " + digitCount);
// Output: Digit count: 6

Did You Know?

Character.isDigit() works with Unicode digits from any language, not just 0-9! For strict 0-9 checking, use c >= '0' && c <= '9'.

Pro Tip

Extract all digits from a string:

java

String text = "Price: $123.45";
String digits = text.replaceAll("[^0-9]", "");
System.out.println(digits); // 12345

Real-World Use Cases

Password validation: Ensure passwords contain numbers
Data extraction: Extract numeric data from text
Input validation: Check if input contains digits
Text analysis: Analyze alphanumeric content

›Short versionOpen if you only have a minute.

Walk the string and count characters that are digits. Do not parse the whole string.
Integer.parseInt("A12B3") throws. The text was never a single number.
Character.isDigit is true for more than '0' through '9', including many Unicode digits.
A leading minus and a decimal point are not digits.

Count digits in text

Count digit characters inside a string.

Working approach

Loop with charAt.
Use c >= '0' && c <= '9' when the task means ASCII digits.
Use Character.isDigit only when other Unicode digits should count.

Walk the sample

Input "A12B3"
A is skipped, 1 and 2 count, B is skipped, 3 counts
The count is 3

The trap

Character.isDigit('\u0663') is true. That character is an Arabic-Indic digit, not the ASCII character 3. A parser that meant 0-9 will over-count.

java

static int asciiDigits(String text) {
    int count = 0;
    for (int i = 0; i < text.length(); i++) {
        char c = text.charAt(i);
        if (c >= '0' && c <= '9') {
            count++;
        }
    }
    return count;
}

Does Character.isDigit count only the characters 0 through 9?

No. It returns true for any character whose Unicode category is a digit, including Arabic-Indic digits.

The range check against '0' and '9' matches the ASCII digits this kind of exercise is usually asking about.

Time O(n), space O(1).

Easy to get wrong

Parsing the whole string with parseInt when the text is mixed.
Assuming isDigit means ASCII.