简单的Java计算器逻辑

Question

我是编程新手，我当时只用1个文本字段制作了一个计算器。 我需要一种方法，该方法可以识别来自这些（+，-，*，/）的哪些字符串字符，以将结果发送到变量，当我单击=按钮时，它将显示结果。 我试图编写类似的内容（1 + 2）并将其保存到变量中，然后当我尝试按=按钮设置变量的文本时，它显示特权错误。

下面是代码

    JButton btnOne = new JButton("1");
    btnOne.addActionListener(new ActionListener() {
        public void actionPerformed(ActionEvent e) {
            sum+=1;
            txtoprtn.setText(txtoprtn.getText()+"1");

    JButton btnTwo = new JButton("2");
    btnTwo.addActionListener(new ActionListener() {
        public void actionPerformed(ActionEvent e) {
            sum+=2;
            txtoprtn.setText(txtoprtn.getText()+"2");
        }
    });

    JButton btnAdd = new JButton("+");
    btnAdd.addActionListener(new ActionListener() {
        public void actionPerformed(ActionEvent e) {
            txtoprtn.setText(txtoprtn.getText()+"+");

        }
    });

    JButton btnEqual = new JButton("=");
    btnEqual.addActionListener(new ActionListener() {
        public void actionPerformed(ActionEvent e) {
            x = Integer.parseInt(txtoprtn.getText());
            txtoprtn.setText(Integer.toString(x));
            }

        }
    );

这是错误

    **Exception in thread "AWT-EventQueue-0" java.lang.NumberFormatException: For input string: "1+2"
at java.lang.NumberFormatException.forInputString(Unknown Source)
at java.lang.Integer.parseInt(Unknown Source)
at java.lang.Integer.parseInt(Unknown Source)
at com.jadv.day01.tasks.AdvCalc$11.actionPerformed(AdvCalc.java:140)
at javax.swing.AbstractButton.fireActionPerformed(Unknown Source)
at javax.swing.AbstractButton$Handler.actionPerformed(Unknown Source)
at javax.swing.DefaultButtonModel.fireActionPerformed(Unknown Source)
at javax.swing.DefaultButtonModel.setPressed(Unknown Source)
at javax.swing.plaf.basic.BasicButtonListener.mouseReleased(Unknown Source)
at java.awt.Component.processMouseEvent(Unknown Source)
at javax.swing.JComponent.processMouseEvent(Unknown Source)
at java.awt.Component.processEvent(Unknown Source)
at java.awt.Container.processEvent(Unknown Source)
at java.awt.Component.dispatchEventImpl(Unknown Source)
at java.awt.Container.dispatchEventImpl(Unknown Source)
at java.awt.Component.dispatchEvent(Unknown Source)
at java.awt.LightweightDispatcher.retargetMouseEvent(Unknown Source)
at java.awt.LightweightDispatcher.processMouseEvent(Unknown Source)
at java.awt.LightweightDispatcher.dispatchEvent(Unknown Source)
at java.awt.Container.dispatchEventImpl(Unknown Source)
at java.awt.Window.dispatchEventImpl(Unknown Source)
at java.awt.Component.dispatchEvent(Unknown Source)
at java.awt.EventQueue.dispatchEventImpl(Unknown Source)
at java.awt.EventQueue.access$500(Unknown Source)
at java.awt.EventQueue$3.run(Unknown Source)
at java.awt.EventQueue$3.run(Unknown Source)
at java.security.AccessController.doPrivileged(Native Method)
at java.security.ProtectionDomain$1.doIntersectionPrivilege(Unknown Source)
at java.security.ProtectionDomain$1.doIntersectionPrivilege(Unknown Source)
at java.awt.EventQueue$4.run(Unknown Source)
at java.awt.EventQueue$4.run(Unknown Source)
at java.security.AccessController.doPrivileged(Native Method)
at java.security.ProtectionDomain$1.doIntersectionPrivilege(Unknown Source)
at java.awt.EventQueue.dispatchEvent(Unknown Source)
at java.awt.EventDispatchThread.pumpOneEventForFilters(Unknown Source)
at java.awt.EventDispatchThread.pumpEventsForFilter(Unknown Source)
at java.awt.EventDispatchThread.pumpEventsForHierarchy(Unknown Source)
at java.awt.EventDispatchThread.pumpEvents(Unknown Source)
at java.awt.EventDispatchThread.pumpEvents(Unknown Source)
at java.awt.EventDispatchThread.run(Unknown Source)**

有什么建议么？？？？

Answer 1

第一个问题

当您尝试只写一个“ 1 + 2”的字符串语句时，您期望发生什么？ 不能仅仅这样评估它。 完成数字插入后，您必须首先对每个＃按钮使用Integer.parseInt() （例如，当您单击+ /-/// *或=时。这意味着数字已完成，您将会说Integer.parseInt("123") instead of Integer.parseInt("1")+Integer.parseInt("2")+Integer.parseInt("3")您要尝试的是解析+将整数解析保留在数字和运算符存储在其他位置的位置（在第二个问题中进行了解释）

第二个问题

当您说“ +”时，它不会解析为任何操作。 您必须将值存储到它们的各个变量中，然后单击“ +”按钮，您将获得正确的答案。 如果要在没有任何外部导入的情况下执行此操作，则必须将操作存储在列表中并评估何时单击=。

btnAdd.addActionListener(new ActionListener() {
        public void actionPerformed(ActionEvent e) {
            operations.add("+");//operations is a list
        }
});

使用导入一次解决两个问题

如果要使用包装进行评估，请使用以下方法：

ScriptEngine evaluationMachine = new ScriptEngineManager().getEngineByName("JavaScript");
engine.eval(foo); //evaluates something like "2+1" into 3.

具有以下进口：

import javax.script.ScriptEngine;
import javax.script.ScriptEngineManager;

编辑：顺便说一句，尽管第二种方法要快得多，但是您可能应该尝试在没有外部导入的情况下编写它，因为您是编程的新手。 在不可避免的情况下，您将不得不考虑该问题。

Answer 2

导致异常的原因是您试图将"1+2"类的字符串解析为该行中的Integer：

x = Integer.parseInt(txtoprtn.getText());

您应该将数字存储在其他位置，即存储在ArrayList中。