Question

嘿所以基本上我有一个简单的聊天机器人的任务，他的程序的目的是让用户输入一个带有JOptionpane的字符串然后程序将搜索用户输入看看是否有什么他们写的包含了我指定的关键词，如果是这样，他们的程序会显示某个消息。到目前为止，iv使用if-else语句编写它，但老师希望我们使用Arrays（我不知道它们是如何工作的，我们只是希望知道）

import javax.swing.JOptionPane;

public class ChatterBot {

public static void main(String args[]) {
String input = "";
String maths = "";
String science = "";
String chemFact = "";
String bioFact = "";
String zooFact = "";
String algFact = "";
String yes = "Well good for you";
String no = "You learn something new everyday :)";
input = JOptionPane
        .showInputDialog("Pick one of the subjects listed to learn a fun fact (english, science, maths) ");

if (input.contains("science")) {
    science = JOptionPane.showInputDialog(
            "What kind of science fact woukd you like to know about? (chem, Biology, Zoology)");
}

else if (input.contains("maths")) {
    maths = JOptionPane.showInputDialog(
            "What kind of maths fact would you like to know about? (algebra, fractions, division) ");
}
if (maths.contains("algebra")) {
    algFact = JOptionPane.showInputDialog(
            "\"Did you know a mathematician who specializes in algebra is called an algebraist? (yes or no)\"");
}
if (algFact.contains("yes")) {
    System.out.println(yes);
} else if (algFact.contains("no")) {
    System.out.println(no);
}

if (science.contains("chem")) {
    chemFact = JOptionPane.showInputDialog(
            "Did you know If you pour a handful of salt into a full glass of water the water level will actually go down rather than overflowing the glass? (yes or no)");
}

if (chemFact.contains("yes")) {
    System.out.println(yes);
} else if (chemFact.contains("no")) {
    System.out.println(no);
}

else if (science.contains("biology")) {
    bioFact = JOptionPane.showInputDialog("Did you know The brain itself cannot feel pain? (yes or no)");
}

if (bioFact.contains("yes")) {
    System.out.println("Well good for you");
} else if (bioFact.contains("no")) {
    System.out.println("You learn something new everyday :)");
}

else if (science.contains("zoology")) {
    zooFact = JOptionPane
            .showInputDialog("Did you know butterflies have taste receptors on their feet? (yes or no)");
}

if (zooFact.contains("yes")) {
    System.out.println("Well good for you");
} else if (zooFact.contains("no")) {
    System.out.println("You learn something new everyday :)");
}
if (input.contains("?")) {
    System.out.println("I will be asking the questions");
}

}

Answer 1

网上有很多关于java数组的好教程。此外，如果您的课程遵循任何类型的教科书，那么它也应该包含数组。 https://www.tutorialspoint.com/java/java_arrays.html 只是从快速的谷歌。

通常，数组是一种数据结构，它将对象保存在类似函数的列表中。

一般来说

type[] var = new type[size];

或

type[] var = {foo0, foo1, foo2...};

真实示例

int[] intergerArray = new int[10];

String[] stringArray = {"Hello", "World"};

索引一般

var =数组变量

index =对象的位置 - 1（计算机从0开始）

var[index]将从index数组

返回位置var中存储的值

索引示例

制作数组：

String[] stringArray = {"This", "is", "my", "first", "array"};

访问第一个值：

stringArray[0];
将值存储在变量中：

String firstWord = stringArray[0];

您甚至可以一次遍历整个数组：

for (int i = 0; i < stringArray.length; i++){
  System.out.print(stringArray[i]);
}

输出： This is my first array

代码

我建议把你可能的输入放在一个数组中（甚至在几个数组中）

String[] subjects = {"English", "Science", "Maths"};

然后，您可以接受来自用户的输入并循环遍历数组，以查看它是否与您支持的输入匹配。此外，通常您希望包含一个＆＃39;默认＆＃39;无效输入的案例。

可能的实施

import javax.swing.JOptionPane;
public class ChatterBot{
  public static void main(String[] args){
    String[] subjects = {"English", "Maths", "Science"};
    String userInput = JOptionPane
      .showInputDialog("Pick one of the subjects listed to learn a fun fact (english, science, maths) ");

    if (userInput.contains(subjects[0]){
     // english facts
    } else if (userInput.contains(subjects[1]){
     // science facts
    } else if (userInput.contains(subjects[2])){
     // maths facts
    }
  } 
}

如何将这个聊天机器人变成一个扫描用户输入的数组

1 个答案: