英文:
Java - Iterate + Instantiate Directory of Classes
问题
我有一个名为Intent的Java类。在一个名为intents的目录中，我定义了几个Intent的子类。现在，在我的运行器类中，我想要实例化每个子类，并将它们放入一个数组列表中，如下所示：
public static String parseTranscript(String transcript) {ArrayList<Intent> intents = new ArrayList<Intent>();File[] intentFiles = new File("./intents").listFiles();for (File fileName : intentFiles) {// 对于在"intents/"中定义的每个Intent，// 创建一个新的类并添加到数组列表中。// intents.add(new fileName.ObjectName());}}
所以，如果我理解正确，我需要打开目录，获取所有文件名，然后根据文件名创建一个对象。怎样是最佳的做法？
文件结构：
- Intent.java- Main.java- intents/- HelloIntent.java- GameIntent.java...
目标是在运行器中不必手动定义每个子类。
英文:
I have a Java class called Intent. In a directory called intents, I have defined several child classes of Intent. Now, in my runner class, I would like to instantiate each of the child classes into an array list as such:
public static String parseTranscript(String transcript) {ArrayList<Intent> intents = new ArrayList<Intent>();File[] intentFiles = new File("./intents").listFiles();for (File fileName : intentFiles) {//for each of the intents defined in "intents/",//create a new class and add to the array list.//intents.add(new fileName.ObjectName());}}
So, if I understand correctly, I would need to open the directory, get all file names, then create an object from that file name. What is the best way to do this?
File structure:
- Intent.java- Main.java- intents/- HelloIntent.java- GameIntent.java...
The goal is to do this without having to manually define each child class in the runner.
答案1
得分: 2
你可以尝试这样做。使用Class.forName，然后检查getSuperclass是否返回Intent。
ArrayList<Intent> intents = new ArrayList<Intent>();String pathName = "./intents";File[] intentFiles = new File(pathName).listFiles();for (File fileName : intentFiles) {if (fileName.isFile() && fileName.getName().endsWith(".class")) {String className = packageName + '.' + fileName.getName().substring(0, fileName.getName().length() - 6);Class<?> aClass = Class.forName(className);if (aClass.getSuperclass().equals(Intent.class)) {Constructor<?> firstConstructor = aClass.getConstructors()[0];Intent o = (Intent) firstConstructor.newInstance(null);intents.add(o);}}}
获取pathName最好这样做：
String pathName = Thread.currentThread().getContextClassLoader().getResources("intents").nextElement().getFile();
获取正确的构造函数可能需要在这里进行更改：
Constructor<?> firstConstructor = aClass.getConstructors()[0];
英文:
You can try in this way. With Class.forName and then checking if getSuperclass returns Intent
ArrayList<Intent> intents = new ArrayList<Intent>();String pathName = "./intents";File[] intentFiles = new File(pathName).listFiles();for (File fileName : intentFiles) {if (fileName.isFile() && fileName.getName().endsWith(".class")) {String className = packageName + '.' + fileName.getName().substring(0, fileName.getName().length() - 6);Class<?> aClass = Class.forName(className);if (aClass.getSuperclass().equals(Intent.class)) {Constructor<?> firstConstructor = aClass.getConstructors()[0];Intent o = (Intent) firstConstructor.newInstance(null);intents.add(o);}}}
Getting pathName would be better do in this way:
String pathName = Thread.currentThread().getContextClassLoader().getResources("intents").nextElement().getFile();
Getting right constructor maybe need change in this:
Constructor<?> firstConstructor = aClass.getConstructors()[0];
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论