Java 在运行时选择可运行 jar 中的主类
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Selecting main class in a runnable jar at runtime
提问by Tomasz
I have two main classes in the app. When I package it to a runnable jar (using Eclipse export function) I have to select a default main class.
我在应用程序中有两个主要类。当我将它打包到一个可运行的 jar（使用 Eclipse 导出功能）时，我必须选择一个默认的主类。
Is there a way to access the non-default main class from the jar at runtime?
有没有办法在运行时从 jar 访问非默认主类？
采纳答案by Jeremy Raymond
You can access both via java -cp myapp.jar com.example.Main1and java -cp myapp.jar com.example.Main2. The default main class in the jar is for when you invoke your app via java -jar myapp.jar.
您可以通过java -cp myapp.jar com.example.Main1和访问java -cp myapp.jar com.example.Main2。jar 中的默认主类用于通过java -jar myapp.jar.
See JAR_(file_format)for more details. When you select the main class in Eclipse this is what gets set in: Main-Class: myPrograms.MyClassinside of the jar manifest META-INF/MANIFEST.MFin side of the jar file.
有关更多详细信息，请参阅JAR_(file_format)。当您在 Eclipse 中选择主类时，这就是设置的内容：Main-Class: myPrograms.MyClass在META-INF/MANIFEST.MFjar 文件一侧的 jar 清单内。
回答by BalusC
Yes, it's possible. You can under each add another class with a main method for that which executes the desired class/method based on the argument.
是的，这是可能的。您可以在每个类下添加另一个具有主要方法的类，该类根据参数执行所需的类/方法。
E.g.
例如
public static void main(String... args) {
if ("foo".equals(args[0])) {
Foo.main(args);
} else if ("bar".equals(args[0])) {
Bar.main(args);
}
}
(don't forget to add the obvious checks yourself such as args.lengthand so on)
（不要忘记自己添加明显的检查，例如args.length等等）
Which you can use as follows:
您可以按如下方式使用：
java -jar YourJar.jar foo
If well designed, this can however make the main()method of the other classes superfluous. E.g.
但是，如果设计得当，这会使main()其他类的方法变得多余。例如
public static void main(String... args) {
if ("foo".equals(args[0])) {
new Foo().execute();
} else if ("bar".equals(args[0])) {
new Bar().execute();
}
}
To abstract this more (to get rid of if/else blocks), you could consider to let them implement some Actioninterface with a void execute()and get hold of them in a Map:
为了进一步抽象（摆脱 if/else 块），您可以考虑让它们Action使用 a实现一些接口并在 a 中void execute()获取它们Map：
private static Map<String, Action> actions = new HashMap<String, Action>();
static {
actions.put("foo", new Foo());
actions.put("bar", new Bar());
}
public static void main(String... args) {
actions.get(args[0]).execute();
}
