英文:
efficient way to check if a group of letters in a string is a variable?
问题
我正在尝试编写一个打印音乐音阶的代码。出于学校的原因，我不能使用列表或数组类型的结构。以下是我目前的代码部分：
public class Musicarum {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
double buildNum = 0.01;
//**************************************************************************************
int a = 1, bb = 2, b = 3, c = 4, db = 4, d = 5, eb = 6, e = 7, f = 8, gb = 9, ab = 10;
//**************************************************************************************
然后在代码的后面部分，一旦你选择要去哪里，它应该打印一个音阶。现在我正在处理的是大调音阶：
if(answer.equals("scales")) {
System.out.println("Would you like to see a major or a minor scale?");
String scale = input.nextLine();
scale = scale.toLowerCase();
if(scale.equals("major"));
System.out.println("What Scale would you like to see?");
scale = input.nextLine();
System.out.println("The scale you would like to see is made of the notes: " + scale.charAt(0));
}}
我目前的问题在于倒数第二行，在打印音阶时。我想知道是否有可能检查字符串scale，并查看内部的单词是否对应预定义的变量。然后，我希望能够使用我分配给变量的数值。我尝试过使用用户输入的第一个字符的 Unicode 值，但我不知道如何将其与我预定义的整数变量匹配。我还尝试过设置类似于 Python 字典的东西，但目前完全不知道如何做。有任何帮助吗？我试图在不使用4000个嵌套的if语句和为每个音阶设置157个变量的情况下完成这个问题。
英文:
I am trying to write a code to print music scales. I cannot use a list or array type of structure (school reasons). here's what i have so far:
public class Musicarum {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
double buildNum = 0.01;
//**************************************************************************************
int a = 1, bb = 2, b = 3, c = 4, db = 4, d = 5, eb = 6, e = 7, f = 8, gb = 9, ab = 10;
//**************************************************************************************
then further in the code it will, once you select where you want to go, it should print a scale. right now i am on major scales:
if(answer.equals("scales")) {
System.out.println("Would you like to see a major or a minor scale?");
String scale = input.nextLine();
scale = scale.toLowerCase();
if(scale.equals("major"));
System.out.println("What Scale would you like to see?");
scale = input.nextLine();
System.out.println("The scale you would like to see is made of the notes: " + scale.charAt(0));
}}
my current problem is in the line above last where i am printing the scale. I am wondering if it is possible to check the string scale and see if the words inside correspond to a variable that i predefined. i then want to be able to use the number value that i assigned to the variable. I have tried using the unicode value of the first character the user types, but i don't know how to use that to match to one of my int variables. I have also looked into setting up something similar to a python dictionary, but I'm completely at a loss for how to do that right now. any help? trying to do this without using 4000 nested if statements and 157 variables for each scale.
答案1
得分: 1
Java并不是一种动态语言，因此将用户字符串“翻译”为变量名虽然可能，但却非常繁琐（也是一个非常高级的主题）。
解决你问题的直接方法是使用 Map 结构（在Java中，映射是类似于Python中字典的概念）。以下是一个快速示例：
Map<String, Integer> scaleByName = new HashMap<>();
scaleByName.put("a", 1);
scaleByName.put("bb", 2);
// ...
Integer value = scaleByName.get(someStringComingFromUserInput);
然后你可以使用类似 scaleByName.get(someStringComingFromUserInput) 的方法，将字符串键映射到对应的整数值。关于如何使用映射，可以在这里找到更多的 javadoc。
英文:
Java isnt a dynamic language, so taking a user string and "translating" that into a variable name is (albeit possible) extremely burdensome (and a very advanced topic).
The straight forward answer to your problem is to use the Map structure (maps are Java's equivalent of python's dicts). A quick example:
Map<String, Integer> scaleByName = new HashMap<>();
scaleByName.put("a", 1);
scaleByName.put("bb", 2); ...
Then you can use calls like scaleByName.get(someStringComingFromUserInput) to map the string key to its Integer value. There is plenty of javadoc that tells you more how to use maps.
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