2

如果用户输入其他字符，我希望用户只输入字母或空格，我想用我搜索过的 jOptionPane 给消息，我尝试了下面的代码

if (!(Pattern.matches("^[a-zA-Z]+$", answerField1.getText())))
        JOptionPane.showMessageDialog(null, "Please enter a valid character", "Error", JOptionPane.ERROR_MESSAGE);

但现在无论我输入什么都会给出错误

现在我更改了代码

Pattern letterPattern = Pattern.compile("^[a-zA-Z]+$");

if (!(letterPattern.matcher(answerField1.getText()).matches()))
      {
    JOptionPane.showMessageDialog(null, "Please enter a valid character", "Error", JOptionPane.ERROR_MESSAGE);
}

现在它只在用户第一次输入数字时给出消息。我该如何解决这个问题

4

3 回答 3

8

使用 a DocumentFilter，这是我做的一个例子，它只接受字母字符和空格：

import javax.swing.JFrame;
import javax.swing.JTextField;
import javax.swing.SwingUtilities;
import javax.swing.text.AbstractDocument;
import javax.swing.text.AttributeSet;
import javax.swing.text.BadLocationException;
import javax.swing.text.DocumentFilter;
import javax.swing.text.DocumentFilter.FilterBypass;

public class Test {

    public Test() {
        initComponents();
    }

    public static void main(String[] args) {
        SwingUtilities.invokeLater(new Runnable() {
            @Override
            public void run() {
                new Test();
            }
        });
    }

    private void initComponents() {
        JFrame frame = new JFrame();
        frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);

        JTextField jtf = new JTextField();
        //add filter to document
        ((AbstractDocument) jtf.getDocument()).setDocumentFilter(new MyDocumentFilter());

        frame.add(jtf);

        frame.pack();
        frame.setVisible(true);
    }
}

class MyDocumentFilter extends DocumentFilter {

    @Override
    public void replace(FilterBypass fb, int i, int i1, String string, AttributeSet as) throws BadLocationException {
        for (int n = string.length(); n > 0; n--) {//an inserted string may be more than a single character i.e a copy and paste of 'aaa123d', also we iterate from the back as super.XX implementation will put last insterted string first and so on thus 'aa123d' would be 'daa', but because we iterate from the back its 'aad' like we want
            char c = string.charAt(n - 1);//get a single character of the string
            System.out.println(c);
            if (Character.isAlphabetic(c) || c == ' ') {//if its an alphabetic character or white space
                super.replace(fb, i, i1, String.valueOf(c), as);//allow update to take place for the given character
            } else {//it was not an alphabetic character or white space
                System.out.println("Not allowed");
            }
        }
    }

    @Override
    public void remove(FilterBypass fb, int i, int i1) throws BadLocationException {
        super.remove(fb, i, i1);
    }

    @Override
    public void insertString(FilterBypass fb, int i, String string, AttributeSet as) throws BadLocationException {
        super.insertString(fb, i, string, as);

    }
}
于 2012-12-27T19:20:11.127 回答
1

第一行末尾有一个分号。所以它并没有真正正确地测试。

像这样的声明（这就是你所拥有的）：

if (condition) ;

如果条件为真，将执行空语句 (;)，然后转到下一行。如果条件为假，它将转到下一行。这两个动作具有相同的结果。

您可以尝试在所有“if”语句上使用大括号。有时很乏味，但更难搞砸。

if (!(Pattern.matches("^[a-zA-Z]+$", answerField1.getText()))) {
    JOptionPane.showMessageDialog(null, "Please enter a valid character", "Error", JOptionPane.ERROR_MESSAGE);
}

这就是我会做的。您可以删除分号。

于 2012-12-27T17:29:08.200 回答
0

此代码可能会帮助您：

answerField1.addKeyListener(new KeyAdapter() {
        @Override
        public void keyPressed(KeyEvent e) {
            char ch = e.getKeyChar();
            if(Character.isDigit(ch)){
                answerField1.setText("");
                JOptionPane.showMessageDialog(null, "Enter Alphabet Only !");
                                     }
                                            }
    });
于 2020-01-17T04:36:16.490 回答