Java - See if a string contains any characters in it

Question

The problem i'm having is when i check to see if the string contains any characters it only looks at the first character not the whole string. For instance I would like to be able to input "123abc" and the characters are recognized so it fails. I also need the string to be 11 characters long and since my program only works with 1 character it cannot go any further.

Here is my code so far:

public static int phoneNumber(int a)
{
   while (invalidinput)
            {
        phoneNumber[a] = myScanner.nextLine();

        if (phoneNumber[a].matches("[0-9+]") && phoneNumber[a].length() == 11 )
        {   
            System.out.println("Continue");
            invalidinput = false;
        }
        else
        {
            System.out.print("Please enter a valid phone number: ");
        }

    }

    return 0;
}

For instance why if i take away the checking to see the phoneNumber.length() it still only registers 1 character so if i enter "12345" it still fails. I can only enter "1" for the program to continue.

If someone could explain how this works to me that would be great

score 5 · Accepted Answer · answered Nov 20 '13 at 16:52

5

Your regex and if condition is wrong. Use it like this:

 if ( phoneNumber[a].matches("^[0-9]{11}$") ) {
    System.out.println("Continue");
    invalidinput = false;
 }

This will only allow phoneNumber[a] to be a 11 character long comprising only digits 0-9

answered Nov 20 '13 at 16:52

anubhava

761,203
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Further reading on boundary matchers for regular expressions in Java: http://docs.oracle.com/javase/tutorial/essential/regex/bounds.html – Klazen108 Nov 20 '13 at 16:57
@Patch: Just click on tick mark on top-left of my answer. Once clicked it will turn green. – anubhava Nov 20 '13 at 17:16
Not a problem, have a nice day. – anubhava Nov 20 '13 at 17:27

score 0 · Answer 2 · answered Nov 20 '13 at 16:52

0

The + should be outside the set, or you could specifically try to match 11 digits like this: ^[0-9]{11}$ (the ^ and $ anchor the match to the start and end of the string).

answered Nov 20 '13 at 16:52

dr01d3k4

91
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score 0 · Answer 3 · answered Nov 20 '13 at 16:54

0

You need to put the "+" after the "]" in your regex. So, you would change it to:

phoneNumber[a].matches("[0-9]+")

answered Nov 20 '13 at 16:54

forgivenson

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score 0 · Answer 4 · answered Nov 20 '13 at 17:00

Why not try using a for loop to go through each character?

Like:

public static int phoneNumber(int a)
{
   while (invalidinput)
            {

       int x = 0;
       for(int i = 0; i < phoneNumber[a].length(); i++)
       {
        char c = phoneNumber[a].charAt(i);
        if(c.matches("[0-9+]")){
        x++;
       }


}
  if (x == phoneNumber[a].length){
    System.out.println("Continue");
    invalidinput = false;

        }
  else
        {
            System.out.print("Please enter a valid phone number: ");
        }

}
return 0;
}

score 0 · Answer 5 · answered Nov 20 '13 at 17:00

Are the legal characters in your phone numbers 0..9 and +? If so, then you should use the regular expression [0-9+]*, which matches zero or more legal characters. (If not, you probably meant [0-9]+.) Also, you can use [0-9+]{11} instead of your explicit check for a length of 11.

The reason that your current code fails, is that String#matches() does not check whether the regular expression matches part of the string, but whether it matches all of the string. You can see this in the JavaDoc, which points you to Matcher#matches(), which "Attempts to match the entire region against the pattern."

Java - See if a string contains any characters in it

5 Answers5

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