如何將控制台遷移到擺動應用程序

Question

我有一個基本的計算程序，已經開始構建GUI，並且有一個窗口。 我的問題是，如何將這兩件事聯系在一起？ 我聽說我應該先制作GUI，但這使我更加困惑。

我想知道如何將后端連接到前端（GUI）

public class Calc_functions {

    //declaring subtraction feild
    public int Sub (int num1, int num2) {
        //returns the value num1 subtract num2
        return num1 - num2;
    }

    //declaring addition field
    public int Add (int fnum, int snum) {
        //returns the value num1 add num2
        return fnum + snum;
    }

    //declaring division field
    public int Div (int fnum, int snum) {
        //returns the value num1 divided by num2
        return fnum / snum;
    }

    //declaring multiplication field
    public int Mult (int fnum, int snum) {
        //returns the value num1 multiplied by num2
        return fnum * snum;
    }

}

---

import java.util.Scanner;

public class calc_main {

    public static void main(String[] args) {
        // calls for the Calc_functions class
        Calc_functions math = new Calc_functions ();
        //waits for user imputs and then store it as a variable
        Scanner numbers = new Scanner(System.in);
        //prints out too interface
        System.out.println("Calulator : Enter two numbers and choose a mathmatic symbol + - x /");
        System.out.println("_____________________");
        //prints out too interface
        System.out.print("First number:");
        int num1 = numbers.nextInt();
        //prints out too interface
        System.out.print("Second number:");
        int num2= numbers.nextInt();
        //prints out too interface
        System.out.print("Enter symbol +  -  x  / of the calculation you would like to perform :");
        String operation= numbers.next();

        // if the user has inputted +, it will carry out the addition of the two variables the user has unputted.
        if (operation.equals("+"))
            System.out.println(math.Add(num1, num2));
        // if the user has inputted -, it will carry out the addition of the two variables the user has unputted.
        else if (operation.equals("-"))
            System.out.println(math.Sub(num1, num2));
        // if the user has inputted x, it will carry out the addition of the two variables the user has unputted.
        else if (operation.equals("x"))
            System.out.println(math.Mult(num1, num2));
        // if the user has inputted /, it will carry out the addition of the two variables the user has unputted.
        else if (operation.equals("/"))
            System.out.println(math.Div(num1, num2));
        else
            System.out.println("The operation is not valid.");

        numbers.close();
        System.exit(0);
    }

}

---

import javax.swing.*;

// some code used from docs.oracle.com
public class Calc_gui {

    private static void GUI(){
        JFrame createshowGUI = new JFrame("Calc_gui");
        createshowGUI.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);

        //Add the ubiquitous "Hello World" label.
        JLabel label = new JLabel("calcgui");
        createshowGUI.getContentPane().add(label);

        //Display the window.
        createshowGUI.pack();
        createshowGUI.setVisible(true);
    }

    public static void main(String[] args) {
        //Schedule a job for the event-dispatching thread:
        //creating and showing this application's GUI.
        javax.swing.SwingUtilities.invokeLater(new Runnable() {
            public void run() {
                GUI();
            }
        });
    }

}

Answer 1

對於這樣一個簡單的任務，您將不需要“后端”和“前端”。 對於此用例 ，只需調用您的計算以及gui組件的相應操作方法即可。 動作方法意味着您向JButton添加例如ActionListener ，然后執行相應的命令，例如執行加法。

然后，您可以將這4種情況下需要執行的代碼提取到偵聽器中，例如（偽代碼，未編譯！）：

void actionPerformed(ActionEvent e)
{    //listener for add-button
    int num1 = Integer.parse(textfield1.getText());
    int num2 = Interger.parse(textfield2.getText());
    textField3.setText(String.valueOf( math.add(num1, num2) ) );
}

...然后通過addActionListener它們連接到按鈕。

上面的代碼從兩個文本字段獲取兩個值，並嘗試將它們轉換為int值。 然后它調用您的計算方法。 有多種方法可以將偵聽器添加到多個按鈕並檢測按下了哪個按鈕（通過將事件的來源與組件進行比較），因此您無需重復所有“從文本字段獲取和設置值”代碼。

這是基本原則。 對於更復雜的應用程序或長時間運行的操作，可能不是這樣做的方式，因為在ActionListener執行它們意味着它們正在阻止EDT，並且不會處理任何GUI事件。