How to validate if a string has 8 characters, at least 1 letter, 1 number and 1 special character?

[Rogerio Kioshi]

I need to validade if a string has 8 characters, at least 1 letter, 1 number and 1 special character. How to write this rule using Regex?

[Jelle Klap]

*cough*

But concerning the regex, what have you come up with so far?

[naved momin]

Rogerio Kioshi wrote:I need to validade if a string has 8 characters, at least 1 letter, 1 number and 1 special character. How to write this rule using Regex?

all you need is this regex

^([a-zA-Z+]+[0-9+]+[&@!#+]+)$

but remember from now one , here we are to help you not to provide ready made solutions so from now first show us what you have tried so far then any one would love to solve your problem . k. have a good day bye, put as much special characters you want in the last square bracket.

[Bear Bibeault]

Just in case it didn't come across, password requirements such as this are ridiculous and make passwords less secure. Who makes up this crap?

The only password restrictions that should ever be in place, in my opinion is a minimum length, and preventing the password from being the same as the username. Anything else is just plain stupid.

[Bear Bibeault]

From the Attorney General of Texas Child Support website:



Stop the insanity!

[Koen Aerts]

And that's why people write down these passwords on a sticky note and hang it on their screen.

[Martin Vajsar]

That's a really, really long way to express "we do not want you to log in to our site".

[Henry Wong]

naved momin wrote:

Rogerio Kioshi wrote:I need to validade if a string has 8 characters, at least 1 letter, 1 number and 1 special character. How to write this rule using Regex?

all you need is this regex

^([a-zA-Z+]+[0-9+]+[&@!#+]+)$

but remember from now one , here we are to help you not to provide ready made solutions so from now first show us what you have tried so far then any one would love to solve your problem . k. have a good day bye, put as much special characters you want in the last square bracket.

A few issues with the regex....

1. The request was for exactly eight characters -- which is not handled by the regex.
2. I don't think that the requirement "at least 1 letter, 1 number and 1 special character", implies that the special character must come after the number, which must come after the letter -- which the regex requires.
3. I don't think that the plus sign within the character class does what you think it does.

Henry

[Jeanne Boyarsky]

Why do you have to use a single regexp? It would be clearer has a bunch of and statements - one for each rule. Some of those rules owuld be a regexp, but a very simple one.

[Campbell Ritchie]

For something that small, would loops and ifs work quicker than a regex?

[Rogerio Kioshi]

Thanks for replying. Actually, I gave up using Regex and made a simple method using the following:

Character.isLetter
Character.isDigit

String strSpecialChars="!@#$%&*()_+=-|<>?{}[]~";
if (strSpecialChars.contains(strChar)){

I think this way is better for code maintenance.

[Michael Parmeley]

You could also take advantage of all the special characters being 33-47 decimal (inclusive) in the ascii table.

public class PasswordStats {
    public void printPasswordStats(String password) {
        int numOfSpecial = 0;
        int numOfLetters = 0;
        int numOfUpperLetters = 0;
        int numOfLowerLetters = 0;
        int numOfDigits = 0;

        byte[] bytes = password.getBytes();
        for (byte tempByte : bytes) {
            if (tempByte >= 33 && tempByte <= 47) {
                numOfSpecial++;
            }

            char tempChar = (char) tempByte;
            if (Character.isDigit(tempChar)) {
                numOfDigits++;
            }

            if (Character.isLetter(tempChar)) {
                numOfLetters++;
            }

            if (Character.isUpperCase(tempChar)) {
                numOfUpperLetters++;
            }

            if (Character.isLowerCase(tempChar)) {
                numOfLowerLetters++;
            }
        }

        System.out.println("numOfSpecial = " + numOfSpecial);
        System.out.println("numOfLetters = " + numOfLetters);
        System.out.println("numOfUpperLetters = " + numOfUpperLetters);
        System.out.println("numOfLowerLetters = " + numOfLowerLetters);
        System.out.println("numOfDigits = " + numOfDigits);
    }

    public static void main(String[] args) {
        new PasswordStats().printPasswordStats("pas$.WOor!/d1%");
    }
}

Results of this code:

numOfSpecial = 5
numOfLetters = 8
numOfUpperLetters = 2
numOfLowerLetters = 6
numOfDigits = 1

[Stuart A. Burkett]

Michael Parmeley wrote:You could also take advantage of all the special characters being 33-47 decimal (inclusive) in the ascii table.

Are you sure ? Without even looking at an ascii table I can tell you it's not true. 33 - 47 covers a range of 15 characters, but Rogerio has 22 characters in his list of special characters.

[Michael Parmeley]

Adrian Burkett wrote:

Michael Parmeley wrote:You could also take advantage of all the special characters being 33-47 decimal (inclusive) in the ascii table.

Are you sure ? Without even looking at an ascii table I can tell you it's not true. 33 - 47 covers a range of 15 characters, but Rogerio has 22 characters in his list of special characters.

Indeed, my mistake. Didn't look at it close enough.

[Henry Wong]

Rogerio Kioshi wrote:Thanks for replying. Actually, I gave up using Regex and made a simple method using the following:

Character.isLetter
Character.isDigit

String strSpecialChars="!@#$%&*()_+=-|<>?{}[]~";
if (strSpecialChars.contains(strChar)){

I think this way is better for code maintenance.

100% agreed. Being able to understand and maintain the code should be of high priority.... However, just in case you are interested in how a regex would look like...

public static boolean validate(String password) {
	return password.matches("(?=.*[a-zA-Z])(?=.*\\d)(?=.*[!@#$%&*()_+=|<>?{}\\[\\]~-]).{8}");
}

Henry

[Winston Gutkowski]

Rogerio Kioshi wrote:Thanks for replying. Actually, I gave up using Regex and made a simple method using the following:

Character.isLetter
Character.isDigit
String strSpecialChars="!@#$%&*()_+=-|<>?{}[]~";
if (strSpecialChars.contains(strChar)){

I think this way is better for code maintenance.

Seems reasonable to me. Just to point out that you could use regexes in a slightly more meaningful way:

Pattern letter = Pattern.compile("[a-zA-z]");
Pattern digit = Pattern.compile("[0-9]");
Pattern special = Pattern.compile ("[!@#$%&*()_+=|<>?{}\\[\\]~-]");
Pattern eight = Pattern.compile (".{8}");
...
public final boolean ok(String password) {
   Matcher hasLetter = letter.matcher(password);
   Matcher hasDigit = digit.matcher(password);
   Matcher hasSpecial = special.matcher(password);
   Matcher hasEight = eight.matcher(password);
   return hasLetter.find() && hasDigit.find() && hasSpecial.find()
      && hasEight.matches();
}

Winston


[Edit]Actually, looking at it:

public final boolean ok(String password) {
   return eight.matcher(password).matches()
      && special.matcher(password).find()
      && digit.matcher(password).find()
      && letter.matcher(password).find();
}

would probably be a lot quicker (with all due deference to Mr. Knuth ).