Getting a name as an input using regex in java

Question

I am a beginner in both, Java and regular expressions. I want to get a name as an input, by which I mean that only names that have English alphabets A-Z, case insensitive and spaces.

I am using a Scanner class to get my input but my code doesn't work. It looks like:

Scanner sc= new Scanner(System.in);
String n;

while(!sc.hasNext("^[a-zA-Z ]*$"))
{
    System.out.println("That's not a name!");
    sc.nextLine();
}
n = sc.next();

I checked my regular expression on the website regex101.com and found out that it works fine.

For example, If I input it my name, Akshay Arora , the regex site says it is okay but my program prints

That's not a name
That's not a name

Same line is printed twice and it again asks me for input. Where am I going wrong?

I am referring to : http://stackoverflow.com/questions/3059333/validating-input-using-java-util-scanner — Akshay Arora, Aug 30 '15 at 12:40
Also your scanner object is `abc1` not `sc`. Also `scanner.hasNext()` returns boolean. — Uma Kanth, Aug 30 '15 at 12:41
@UmaKanth, the third example of vowels uses regular expressions this way. — Akshay Arora, Aug 30 '15 at 12:41
I have a scanner object `sc` defined at the top, just my mistake I left the `abc1`. Should delete that. — Akshay Arora, Aug 30 '15 at 12:41

score 2 · Accepted Answer · answered Aug 30 '15 at 12:45

2

Two parts are wrong:

$ and ^ anchors are considered in the context of entire input, not in the context of the next token. It will never match, unless the input has a single line that matches the pattern in its entirety.
You use default delimiters, which include spaces; therefore, Scanner will never return a token with a space in it.

Here is how you can fix this:

Scanner sc = new Scanner(System.in);
sc.useDelimiter("\n");
String n;

while(!sc.hasNext("[a-zA-Z ]+"))
{
    System.out.println("That's not a name!");
    sc.nextLine();
}
n = sc.next();

Demo.

answered Aug 30 '15 at 12:45

Sergey Kalinichenko

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Great. This answers the question in the context in which it was asked. – Akshay Arora Aug 30 '15 at 12:50
1

Worth mentioning that this pattern allows arbitrary number of spaces which is not suitable for name checks – Aleksandr Erokhin Aug 30 '15 at 13:16

score 1 · Answer 2 · answered Aug 30 '15 at 12:45

Here its sample program related to regex.

public class Program {

    public static void main(String[] args) {

    Scanner sc = new Scanner(System.in);

    String inputName = sc.next();

    String regex = "^[a-zA-Z ]*$";
    // Compile this pattern.
    Pattern pattern = Pattern.compile(regex);

    // See if this String matches.
    Matcher m = pattern.matcher(inputName);
    if (m.matches()) {
        System.out.println("Valid Name");
    } else
        System.out.println("Invalid Name");

    }
}

Hope this will help you

Getting a name as an input using regex in java

2 Answers2