In Java, how to find if first character in a strin

问题:

In Java, find if the first character in a string is upper case without using regular expressions.

回答1:

Assuming s is non-empty:

Character.isUpperCase(s.charAt(0))

or, as mentioned by divec, to make it work for characters with code points above U+FFFF:

Character.isUpperCase(s.codePointAt(0));

回答2:

Actually, this is subtler than it looks.

The code above would give the incorrect answer for a lower case character whose code point was above U+FFFF (such as U+1D4C3, MATHEMATICAL SCRIPT SMALL N). String.charAt would return a UTF-16 surrogate pair, which is not a character, but rather half the character, so to speak. So you have to use String.codePointAt, which returns an int above 0xFFFF (not a char). You would do:

Character.isUpperCase(s.codePointAt(0));

Don't feel bad overlooked this; almost all Java coders handle UTF-16 badly, because the terminology misleadingly makes you think that each "char" value represents a character. UTF-16 sucks, because it is almost fixed width but not quite. So non-fixed-width edge cases tend not to get tested. Until one day, some document comes in which contains a character like U+1D4C3, and your entire system blows up.

回答3:

There is many ways to do that, but the simplest seems to be the following one:

boolean isUpperCase = Character.isUpperCase("My String".charAt(0));

回答4:

we can find upper case letter by using regular expression as well

private static void findUppercaseFirstLetterInString(String content) {
    Matcher m = Pattern
            .compile("([a-z])([a-z]*)", Pattern.CASE_INSENSITIVE).matcher(
                    content);
    System.out.println("Given input string : " + content);
    while (m.find()) {
        if (m.group(1).equals(m.group(1).toUpperCase())) {
            System.out.println("First Letter Upper case match found :"
                    + m.group());
        }
    }
}

for detailed example . please visit http://www.onlinecodegeek.com/2015/09/how-to-determines-if-string-starts-with.html

回答5:

Don't forget to check whether the string is empty or null. If we forget checking null or empty then we would get NullPointerException or StringIndexOutOfBoundException if a given String is null or empty.

public class StartWithUpperCase{

        public static void main(String[] args){

            String str1 = ""; //StringIndexOfBoundException if 
                              //empty checking not handled
            String str2 = null; //NullPointerException if 
                                //null checking is not handled.
            String str3 = "Starts with upper case";
            String str4 = "starts with lower case";

            System.out.println(startWithUpperCase(str1)); //false
            System.out.println(startWithUpperCase(str2)); //false
            System.out.println(startWithUpperCase(str3)); //true
            System.out.println(startWithUpperCase(str4)); //false



        }

        public static boolean startWithUpperCase(String givenString){

            if(null == givenString || givenString.isEmpty() ) return false;
            else return (Character.isUpperCase( givenString.codePointAt(0) ) );
        }

    }

回答6:

Make sure you first check for null and empty and ten converts existing string to upper. Use S.O.P if want to see outputs otherwise boolean like Rabiz did.

 public static void main(String[] args)
 {
     System.out.println("Enter name");
     Scanner kb = new Scanner (System.in);
     String text =  kb.next();

     if ( null == text || text.isEmpty())
     {
         System.out.println("Text empty");
     }
     else if (text.charAt(0) == (text.toUpperCase().charAt(0)))
     {
         System.out.println("First letter in word "+ text + " is upper case");
     }
  }

回答7:

If you have to check it out manually you can do int a = s.charAt(0)

If the value of a is between 65 to 90 it is upper case.

回答8:

String yourString = "yadayada";
if (Character.isUpperCase(yourString.charAt(0))) {
    // print something
} else {
    // print something else
}