沒有數字按鈕的GUI計算器

Question

您好，我有這個計算器，我希望它是無按鈕的，就像在文本字段中輸入用戶輸入一樣，但是我不知道如何使它像這樣工作。 我想這樣簡化它，但是我不知道如何在沒有數字按鈕的情況下使它工作。

import java.awt.*;
import javax.swing.*;
import java.awt.event.*;

class MP2_2 extends JFrame implements ActionListener {

Container c;
JTextField result;
JPanel p = new JPanel();
JButton b[] = new JButton[16];
String s[] = {"0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "+", "-", "/", "*", "=", "C"};
//----------------------Me------------------------
String Screen = "", monitor1 = "", monitor2 = "", OperationOnScreen = "";
boolean CommandEmpty = true, switcher = true;
double R = Integer.MIN_VALUE, L = Integer.MIN_VALUE;

//------------------------------------------------

public MP2_2() {
    super("MP2_2");
    c = getContentPane();
    result = new JTextField();
    result.setEditable(true);

    p.setLayout(new GridLayout(4, 4));
    for (int i = 0; i < 16; i++) {
        b[i] = new JButton(s[i]);
        b[i].addActionListener(this);
        p.add(b[i]);
    }
    c.add(result, BorderLayout.NORTH);
    c.add(p);

}//End Constructor

public static void main(String[] args) {
    MP2_2 calcu = new MP2_2();
    calcu.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
    calcu.setSize(300, 300);
    calcu.setVisible(true);
}

//------------------------------------------------              

public void actionPerformed(ActionEvent event) {
    for (int i = 0; i <= 9; i++)//Numbers
    {
        if (event.getSource() == b[i]) {
            Screen += i;
            result.setText("");
            result.setText(Screen);
        }
    }

    for (int i = 10; i <= 14; i++)//Commands
    {
        if (event.getSource() == b[i]) {
            if (result.getText().lastIndexOf(OperationOnScreen) != -1)//prevent exception
            {
                result.setText(result.getText().substring(0, result.getText().lastIndexOf(OperationOnScreen)) + s[i]);
            } else {
                result.setText(result.getText() + s[i]);
            }
            OperationOnScreen = s[i];

            if (switcher) {
                monitor1 = s[i];
                switcher = false;
            } else {
                monitor2 = s[i];
                switcher = true;
            }

            if (monitor1 != monitor2 && monitor2 != "") {
                if (switcher) //execute older,send sign newer
                {
                    Calc(event, monitor1.charAt(0), monitor2);
                } else {
                    Calc(event, monitor2.charAt(0), monitor1);
                }
            }
            if (s[i] != "=") //calc returns 0
            {
                Calc(event, s[i].charAt(0), s[i]);
            }
        }
    }

    if (event.getSource() == b[15]) //Clear
    {
        Screen = "";
        monitor1 = "";
        monitor2 = "";
        switcher = true;
        CommandEmpty = true;
        result.setText("");
    }
}//end actionPerformed

public void Calc(ActionEvent event, char OpType, String Operator) {
    if (Operator == "=") {
        Operator = "";
    }

    if (CommandEmpty && Screen == "") {
        return;
    } else if (CommandEmpty && Screen != "") {
        R = Integer.parseInt(Screen);
        result.setText(Screen + Operator);
        Screen = "";
        CommandEmpty = false;
    } else if (!CommandEmpty && Screen != "") {
        L = Integer.parseInt(Screen);
        R = Operations(R, L, OpType);//calculate
        Screen = "";
        result.setText("");
        result.setText(R + Operator);
    }
}//End Calc             

public static double Operations(double R, double L, char op) {
    switch (op) {
        case '+':
            return R + L;
        case '-':
            return R - L;
        case '*':
            return R * L;
        case '/':
            return R / L;
    }
    return 0;
}
}//end class

編輯：我想我的問題不是很抱歉。 我希望我的程序也支持鍵盤輸入，該怎么辦？ 因此，我最終可以刪除圖片中的數字按鈕。

Answer 1

刪除按鈕時，需要通過調用result.getText（）來恢復按鈕的操作。

這是適用於您的數字按鈕的代碼。 它設置屏幕的值。

for (int i = 0; i <= 9; i++)//Numbers
{
    if (event.getSource() == b[i]) {
        Screen += i;
        result.setText("");
        result.setText(Screen);
    }
}

因此，當您按下操作按鈕時，您需要設置屏幕。 這是您的命令循環。 我添加了Screen = result.getText（）; 在else塊中。 它為我工作。

for (int i = 10; i <= 14; i++)//Commands
{
    if (event.getSource() == b[i]) {
        if (result.getText().lastIndexOf(OperationOnScreen) != -1)//prevent exception
        {
            result.setText(result.getText().substring(0, result.getText().lastIndexOf(OperationOnScreen)) + s[i]);
        } else {
            Screen = result.getText();
            result.setText(result.getText() + s[i]);
        }