Regular expression for alphanumeric

Question

I want a regular expression in java that must contain at least a alphabet and a number at any position. It is for the password that contain the digits as well as numbers.

This should work for:

"1a1b23nh" Accepted

"bc112w" Accepted

"abc" Not accepted

"123" Not accepted

Special characters are not allowed.

Answer 1

([0-9]+[a-zA-Z][0-9a-zA-Z]*)|([a-zA-Z]+[0-9][0-9a-zA-Z]*)

Answer 2

(([a-z]+[0-9]+)+|(([0-9]+[a-z]+)+))[0-9a-z]*

---

How about a simple content check? Check if there are number(s) and character(s)

String input = "b45z4d";
boolean alpha = false;
boolean numeric = false;
boolean accepted = true;
for (int i = 0; i < input.length(); ++i)
{
    char c = input.charAt(i);
    if (Character.isDigit(c))
    {
        numeric = true;
    } else if (Character.isLetter(c))
    {
        alpha = true;
    } else
    {
        accepted = false;
        break;
    }
}

if (accepted && alpha && numeric)
{
    // Then it is correct
}

Answer 3

I know that the question already have been answered, and accepted, but this is what I would do:

Pattern pattern = Pattern.compile("(?i)(?:((?:\\d+[a-z]+)|(?:[a-z]+\\d+))\\w*)");

Object[][] tests = new Object[][] {
        { "1a1b23nh", Boolean.valueOf(true) },
        { "bc112w", Boolean.valueOf(true) },
        { "abc", Boolean.valueOf(false) },
        { "123", Boolean.valueOf(false) }
};

for (Object[] test : tests) {
    boolean result = pattern.matcher((String)test[0]).matches();
    boolean expected = ((Boolean)test[1]).booleanValue();
    System.out.print(test[0] + (result ? "\t " : "\t not ") + "accepted"); 
    System.out.println(result != expected ? "\t test failed" : "");
}
System.out.println("\nAll checks have been executed");

(?i) makes the regexp case insensitive.