如何只使用java正則表達式匹配字母，匹配方法？

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]");
    }
}

輸出為False。 我哪里錯了？ 此外，我想檢查一個單詞是否只包含字母，並且可能或可能不以一個句點結束。 這是什么樣的正則表達式？

即“abc”“abc。” 有效但“abc ..”無效。

我可以使用indexOf()方法來解決它，但我想知道是否可以使用單個正則表達式。

Answer 1

"[a-zA-Z]"只匹配一個字符。 要匹配多個字符，請使用"[a-zA-Z]+" 。

由於點是任何角色的小丑，你必須掩蓋它： "abc\\." 要使點可選，您需要一個問號： "abc\\.?"

如果在代碼中將Pattern寫為文字常量，則必須屏蔽反斜杠：

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

結合兩種模式：

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

而不是a-zA-Z，\\ w通常更合適，因為它捕獲äöüßø等外來字符：

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

Answer 2

[A-Za-z ]*匹配字母和空格。

Answer 3

matches方法執行匹配的整行，即它等同於find()和'^ abc $'。 所以，只需使用Pattern.compile("[a-zA-Z]").matcher(str).find()代替。 然后修復你的正則表達式。 正如@user未知提到的，你的正則表達式實際上只匹配一個字符。 你應該說[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

這里有三個問題：

1. 只需使用String.matches() - 如果API在那里，請使用它
2. 在java中，“匹配”意味着“匹配整個輸入”，恕我直言，這是反直覺的，所以讓你的方法的API通過讓調用者考慮匹配部分輸入來反映這一點，如你的例子所示
3. 你的正則表達式只匹配1個字符

我建議你使用這樣的代碼：

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
}