How to match letters only using java regex, matches method?

Question

import java.util.regex.Pattern;

class HowEasy {
    public boolean matches(String regex) {
        System.out.println(Pattern.matches(regex, "abcABC   "));
        return Pattern.matches(regex, "abcABC");
    }

    public static void main(String[] args) {
        HowEasy words = new HowEasy();
        words.matches("[a-zA-Z]");
    }
}

The output is False. Where am I going wrong? Also I want to check if a word contains only letters and may or maynot end with a single period. What is the regex for that?

ie "abc" "abc." is valid but "abc.." is not valid.

I can use indexOf() method to solve it, but I want to know if it is possible to use a single regex.

Answer 1

"[a-zA-Z]" matches only one character. To match multiple characters, use "[a-zA-Z]+" .

Since a dot is a joker for any character, you have to mask it: "abc\\." To make the dot optional, you need a question mark: "abc\\.?"

If you write the Pattern as literal constant in your code, you have to mask the backslash:

System.out.println ("abc".matches ("abc\\.?"));
System.out.println ("abc.".matches ("abc\\.?"));
System.out.println ("abc..".matches ("abc\\.?"));

Combining both patterns:

System.out.println ("abc.".matches ("[a-zA-Z]+\\.?"));

Instead of a-zA-Z, \\w is often more appropriate, since it captures foreign characters like äöüßø and so on:

System.out.println ("abc.".matches ("\\w+\\.?"));   

Answer 2

[A-Za-z ]*匹配字母和空格。

Answer 3

matches method performs matching of full line, ie it is equivalent to find() with '^abc$'. So, just use Pattern.compile("[a-zA-Z]").matcher(str).find() instead. Then fix your regex. As @user unknown mentioned your regex actually matches only one character. You probably should say [a-zA-Z]+

Answer 4

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.regex.*;

/* Write an application that prompts the user for a String that contains at least
 * five letters and at least five digits. Continuously re-prompt the user until a
 * valid String is entered. Display a message indicating whether the user was
 * successful or did not enter enough digits, letters, or both.
 */
public class FiveLettersAndDigits {

  private static String readIn() { // read input from stdin
    StringBuilder sb = new StringBuilder();
    int c = 0;
    try { // do not use try-with-resources. We don't want to close the stdin stream
      BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
      while ((c = reader.read()) != 0) { // read all characters until null
        // We don't want new lines, although we must consume them.
        if (c != 13 && c != 10) {
          sb.append((char) c);
        } else {
          break; // break on new line (or else the loop won't terminate)
        }
      }
      // reader.readLine(); // get the trailing new line
    } catch (IOException ex) {
      System.err.println("Failed to read user input!");
      ex.printStackTrace(System.err);
    }

    return sb.toString().trim();
  }

  /**
   * Check the given input against a pattern
   *
   * @return the number of matches
   */
  private static int getitemCount(String input, String pattern) {
    int count = 0;

    try {
      Pattern p = Pattern.compile(pattern);
      Matcher m = p.matcher(input);
      while (m.find()) { // count the number of times the pattern matches
        count++;
      }
    } catch (PatternSyntaxException ex) {
      System.err.println("Failed to test input String \"" + input + "\" for matches to pattern \"" + pattern + "\"!");
      ex.printStackTrace(System.err);
    }

    return count;
  }

  private static String reprompt() {
    System.out.print("Entered input is invalid! Please enter five letters and five digits in any order: ");

    String in = readIn();

    return in;
  }

  public static void main(String[] args) {
    int letters = 0, digits = 0;
    String in = null;
    System.out.print("Please enter five letters and five digits in any order: ");
    in = readIn();
    while (letters < 5 || digits < 5) { // will keep occuring until the user enters sufficient input
      if (null != in && in.length() > 9) { // must be at least 10 chars long in order to contain both
        // count the letters and numbers. If there are enough, this loop won't happen again.
        letters = getitemCount(in, "[A-Za-z]");
        digits = getitemCount(in, "[0-9]");

        if (letters < 5 || digits < 5) {
          in = reprompt(); // reset in case we need to go around again.
        }
      } else {
        in = reprompt();
      }
    }
  }

}

Answer 5

Three problems here:

1. Just use String.matches() - if the API is there, use it
2. In java "matches" means "matches the entire input", which IMHO is counter-intuitive, so let your method's API reflect that by letting callers think about matching part of the input as your example suggests
3. You regex matches only 1 character

I recommend you use code like this:

public boolean matches(String regex) {
    regex = "^.*" + regex + ".*$"; // pad with regex to allow partial matching
    System.out.println("abcABC   ".matches(regex));
    return "abcABC   ".matches(regex);
}

public static void main(String[] args) {
    HowEasy words = new HowEasy();
    words.matches("[a-zA-Z]+"); // added "+" (ie 1-to-n of) to character class
}