正则表达式允许使用Java中的特定特殊字符

Question

I need to know the regular expression for string that contains alphanumeric characters, @, underscore(_), full stop(.)and not any blank spaces. And also for alphanumeric characters and it allow spaces. I tried with this regex,

^[_A-Za-z0-9-\\.\\@]$ and ^[A-Za-z0-9-\\s]$

CODE:

private static final String Username_REGEX ="^[_A-Za-z0-9.@-]$";

public static boolean isUsername(EditText editText, boolean required) {
    return isValid(editText, Username_REGEX,Username_MSG, required);
}


public static boolean isValid(EditText editText, String regex, String errMsg, boolean required) {
    String text = editText.getText().toString().trim();

    editText.setError(null);

  if ( required && !hasTextemt(editText) ) return false;

    if (required && !Pattern.matches(regex, text)) {
        editText.setError(errMsg);
        return false;
    };

    return true;
}

public static boolean hasTextemt(EditText editText) {

    String text = editText.getText().toString().trim();
    editText.setError(null);


    if (text.length() == 0) {
        editText.setError(emt);
        return false;
    }

    return true;
}

Is this correct? I did not get proper result. Can anyone guide me?

Answer 1

Move the dash - at the end of the character class:

^[_A-Za-z0-9.@-]+$

and

^[A-Za-z0-9\\s-]+$

Between two characters it means a range.

Edit: You also need a + modifier to match one or more of the characters in the character class.

Answer 2

I am assuming that you are getting this input via an EditText widget. So inside the layout of the XML file you can add the following properties by which it will receive only specified characters. :

android:digits="abcdefghijklmnopqrstuvwxyz0123456789,.-@_"

note that it wont allow any capital letter.

just add any digits/keys you want your user to be able to enter. If you are not worried about the patterns and number of occurrence of any character then you don't even need any regex .

Hope it helps

Answer 3

Try

"[\\w@\\.]+" //for alphanumeric, @, .

"[\\w\\s]+" //for alphanumeric, spaces

Add ^ and $ if you need that matches the whole word.

PS: For testing regexp I always use RegexPlanet (not spam :P)

Hope it helps.

Answer 4

You are only missing a quantifier . In your expression ^[_A-Za-z0-9.@-]$ , the character class [_A-Za-z0-9.@-] matches exactly one character out of the class. To allow repeated characters, you need to define a quantifier .

* short for {0,} matches 0 or more characters (==> this allows the empty string!)

+ short for {1,} matches 1 or more characters

{n,m} matches minimum n and maximum m characters.

So your regex would look like

^[_A-Za-z0-9.@-]+$

if you require 1 or more characters, or

^[_A-Za-z0-9.@-]{6,20}$

if you want at least 6 characters and at most 20.

Other things:

• You can replace _A-Za-z0-9 by \\w , but be aware, \\w is Unicode based and contains all letters and digits from all languages.

• A-Za-z is only ASCII, maybe you want to have a look at Unicode properties . With eg \\p{L} you can match a letter of any language.

Answer 5

You're missing a plus sign (meaning one or more) at the end of the character class, and you can simplify considerably:

^[\\w.@]+$

Characters within a character class lose their special meanings so don't need to be escaped, except for square brackets and a couple of others.

---

For alphanumeric and spaces only, that is only combinations of letters, numbers and spaces:

^[a-zA-Z0-9 ]+$