我正在编写一个 Java 程序，该程序旨在根据用户的输入与用户一起播放 20 个问题的版本
为此，我使用了很多嵌套的 if/else 语句。
我有三个主要分支（活的动物、活的植物、非生物），每个分支都有多个分支。做出60个不同的决定。
我很难让它合作并控制所有 if/else 语句。由于必须重新启动这么多，我还没有太多代码，但目前我在：
System.out.println("Think of Something");
System.out.println("Is it a living animal, living plant, or non-living thing? ");
String user = get.nextLine();
if (user.equals("living animal")); {
//starts animal tree
System.out.println("Does it have feathers, fur, or neither?");
String user2 = get.nextLine();
if (user2.equals("feathers")); {
System.out.println("is it bigger than a soccer ball?");
}
} else if (user2.equals("fur")); {
System.out.println("is it domesticated?");
// end animal tree
} else if (user.equals("living plant")); {
// start plant tree
System.out.println("is it a tree?");
}
} // end method
} //end program
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烙印99
您正在使用以下语法写出您的if语句：if (user2.equals("feathers"));{ System.out.println("is it bigger than a soccer ball?");}但是，if块的主体将始终执行，因为您有一个分号过早地完成了语句：if (user2.equals("feathers")); // <-- The semicolon here finishes the if statement, which means the if statement does nothing{ System.out.println("is it bigger than a soccer ball?"); // <-- This line is ran no matter what the if statement was}if基本上，要使orelse if语句正常工作，您所要做的就是删除不需要的分号。 -
至尊宝的传说
作为一个例子，说明如何解决一个难以处理的问题。开箱即用的程序并不重要。当“很难让它合作并控制所有 if/else 语句”时，它回答了如何简化事情的问题。如果您愿意，可以为这种情况提供一种策略。我也做了一些演示。在实践中，你会做看起来很方便的事情。另外：为了简单起见，我将所有内容都设置为静态 - 在成长的应用程序中，您肯定会使用实例。第 1 步：您从一个非常简单的类框架开始。简单是关键。不要投入太多。只需画出你知道你想要它做什么：public class TwentyQuestions{ static void mainQuestioning(){ System.out.println("Is it a living animal, living plant, or non-living thing? "); String user = get.nextLine(); switch(user){ case "living animal" : askLivingAnimalQuestions(); break; case "living plant": askLivingPlantQuestions(); break; case "non-living": askNoneLivingQuestions(); break; default: handleWrongInput(); } } }当然，上面的那个东西不会编译，因为细节现在还没有实现（缺少一些方法） - 但请注意问题是如何简化很多的（没有嵌套的 if），很可能你可以立即看到它应该是什么做。保持简单，直截了当是关键。第 2 步：现在您可以轻松地创建到目前为止所绘制的方法。让我们这样做：public class TwentyQuestions{ static void handleWrongInput(){ System.err.println("I am no longer playing with you as you don't answer my question properly"); System.exit(1); } static void askLivingAnimalQuestions(){ System.out.println("Does it have feathers, fur, or neither?"); String user = get.nextLine(); switch(user){ case "feathers": askLivinAnimalWithFeathersQuestions(); break; case "fur": askLivinAnimalWithFurQuestions(); break; default: handleWrongInput(); } } static void askLivingPlantQuestions(){ System.out.println("is it a tree?"); String user = get.nextLine(); if("yes".equals(user)){ System.out.println("So its a tree!"); return; } } static void askNoneLivingQuestions(){ System.out.println("WhateverNoneLivingQuestion ?"); String user = get.nextLine(); switch(user){ //add possible responses here. default: handleWrongInput(); } } static void mainQuestioning(){ System.out.println("Is it a living animal, living plant, or non-living thing? "); String user = get.nextLine(); switch(user){ case "living animal" : askLivingAnimalQuestions(); break; case "living plant": askLivingPlantQuestions(); break; case "non-living": askNoneLivingQuestions(); break; default: handleWrongInput(); } } }现在我把问题分解得更厉害了。但它仍然/再次无法编译，因为缺少用于有毛皮的动物和有羽毛的动物的方法。第 3 步：也实施它们：public class TwentyQuestions{ static void handleWrongInput(){ System.err.println("I am no longer playing with you if you don't answer my question properly"); System.exit(1); } static void askLivinAnimalWithFeathersQuestions(){ System.out.println("is it bigger than a soccer ball?"); String user = get.nextLine(); //don't know how you want to continue; //.... } static void askLivinAnimalWithFurQuestions(){ System.out.println("is it domesticated?"); String user = get.nextLine(); //don't know how you want to continue; //.... } static void askLivingAnimalQuestions(){ System.out.println("Does it have feathers, fur, or neither?"); String user = get.nextLine(); switch(user){ case "feathers": askLivinAnimalWithFeathersQuestions(); break; case "fur": askLivinAnimalWithFurQuestions(); break; default: handleWrongInput(); } } static void askLivingPlantQuestions(){ System.out.println("is it a tree?"); String user = get.nextLine(); if("yes".equals(user)){ System.out.println("So its a tree!"); return; } } static void askNoneLivingQuestions(){ System.out.println("WhateverNoneLivingQuestion ?"); String user = get.nextLine(); switch(user){ //add possible responses here. default: handleWrongInput(); } } static void mainQuestioning(){ System.out.println("Is it a living animal, living plant, or non-living thing? "); String user = get.nextLine(); switch(user){ case "living animal" : askLivingAnimalQuestions(); break; case "living plant": askLivingPlantQuestions(); break; case "non-living": askNoneLivingQuestions(); break; default: handleWrongInput(); } } }请注意所有导致您麻烦的嵌套 if/else 是如何消失的。完成：现在，如果您另外实现缺少的问题并添加一个在 main(String[] args) 中初始化的扫描仪“get”，那么您应该在那里。现在应该很容易了。嗯.. 这可能为您提供了很多方法来解决 20 个嵌套问题：这是由于您拥有的可能性的数量。你必须处理那么多的问题和答案。没办法绕过去。最好让他们干净地呆在自己专用的地方，而不是在某个地方闲逛（你整理好并将所有东西放在它的位置 - 你必须处理的案件/问题的数量保持不变）。然而，在一个成熟的应用程序中，您可以将所有问题和答案放在一个像树一样的数据结构中。这样你就可以避免大量的方法，并有一些通用的方法，而不是只走树....[您也可以为您需要但尚未实现的东西创建什么都不做的临时方法（“存根”），以便在您仍在开发时使其编译。]这是一个完整类的示例，它编译并执行到目前为止的问题：import java.util.Scanner;/** * * @author Kai */public class TwentyQuestions { static Scanner get = new Scanner(System.in); static void handleWrongInput() { System.err.println("I am no longer playing with you if you don't answer my question properly"); System.exit(1); } static void askLivinAnimalWithFeathersQuestions() { System.out.println("is it bigger than a soccer ball?"); String user = get.nextLine(); //don't know how you want to continue; //.... } static void askLivinAnimalWithFurQuestions() { System.out.println("is it domesticated?"); String user = get.nextLine(); //don't know how you want to continue; //.... } static void askLivingAnimalQuestions() { System.out.println("Does it have feathers, fur, or neither?"); String user = get.nextLine(); switch (user) { case "feathers": askLivinAnimalWithFeathersQuestions(); break; case "fur": askLivinAnimalWithFurQuestions(); break; default: handleWrongInput(); } } static void askLivingPlantQuestions() { System.out.println("is it a tree?"); String user = get.nextLine(); if ("yes".equals(user)) { System.out.println("So its a tree!"); return; } } static void askNoneLivingQuestions() { System.out.println("WhateverNoneLivingQuestion ?"); String user = get.nextLine(); switch (user) { //add possible responses here. default: handleWrongInput(); } } static void mainQuestioning() { System.out.println("Is it a living animal, living plant, or non-living thing? "); String user = get.nextLine(); switch (user) { case "living animal": askLivingAnimalQuestions(); break; case "living plant": askLivingPlantQuestions(); break; case "non-living": askNoneLivingQuestions(); break; default: handleWrongInput(); } } /** * @param args the command line arguments */ public static void main(String[] args) { mainQuestioning(); }}示例运行：Is it a living animal, living plant, or non-living thing? living animalDoes it have feathers, fur, or neither?furis it domesticated?yesBUILD SUCCESSFUL (total time: 30 seconds) -
繁花如伊
在这种情况下，您没有正确使用 if else 可能是正确的缩进和注释也可以帮助您。System.out.println("Think of Something");System.out.println("Is it a living animal, living plant, or non-living thing? ");String user = get.nextLine();// start methodif (user.equals("living animal")); { //starts animal tree System.out.println("Does it have feathers, fur, or neither?"); String user2 = get.nextLine(); if (user2.equals("feathers")); { System.out.println("is it bigger than a soccer ball?"); } else if (user2.equals("fur")); { System.out.println("is it domesticated?"); }} else if (user.equals("living plant")); { //starts plant tree System.out.println("is it a tree?");} // end method
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