Java ProcessBuilder - 启动另一个进程/JVM - HowTo？
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ProcessBuilder - Start another process / JVM - HowTo?
提问by Stefan K.
I'm writing a network app, where each Client has a Singleton ClientManager. For testing, I would like to create several clients (each in their own VM / process) without starting the program by hand n-times.
我正在编写一个网络应用程序，其中每个客户端都有一个单例 ClientManager。为了测试，我想创建多个客户端（每个客户端都在自己的 VM/进程中），而无需手动启动程序 n 次。
The following two questions on stackoverflow already describe how-to do that:
stackoverflow 上的以下两个问题已经描述了如何做到这一点：
- Is thisreally the best way to start a second JVM from Java code?
- Java: Executing a Java application in a separate process
My Code is based on these, but it's not working:
我的代码基于这些，但它不起作用：
- The main program doesn't continue after spawn is called.
- The spawned code doesn't get executed.
- 调用 spawn 后主程序不会继续。
- 生成的代码不会被执行。
Here's the complete code using ProcessBuilder:
这是使用ProcessBuilder的完整代码：
public class NewVM {
static class HelloWorld2 {
public static void main(String[] args) {
System.out.println("Hello World");
System.err.println("Hello World 2");
}
}
public static void main(String[] args) throws Exception {
startSecondJVM(HelloWorld2.class, true);
startSecondJVM(HelloWorld2.class, false);
System.out.println("Main");
}
public static void startSecondJVM(Class<? extends Object> clazz, boolean redirectStream) throws Exception {
System.out.println(clazz.getCanonicalName());
String separator = System.getProperty("file.separator");
String classpath = System.getProperty("java.class.path");
String path = System.getProperty("java.home")
+ separator + "bin" + separator + "java";
ProcessBuilder processBuilder =
new ProcessBuilder(path, "-cp",
classpath,
clazz.getCanonicalName());
processBuilder.redirectErrorStream(redirectStream);
Process process = processBuilder.start();
process.waitFor();
System.out.println("Fin");
}
}
What am I doing wrong???
我究竟做错了什么？？？
Btw:
顺便提一句：
- I'm using Eclipse.
- The Singleton problem is a simplifiedexample. Please do notsuggest creating a factory.
- 我正在使用 Eclipse。
- 单例问题是一个简化的例子。请不要不建议创建一个工厂。
Solution:HelloWorld2 mustn't be an inner class.
解决方案：HelloWorld2 不能是内部类。
采纳答案by emory
I suggest you make HelloWorld2 a top level class. It appears java expects a top level class.
我建议您将 HelloWorld2 设为顶级类。看来java需要一个顶级类。
This is the code I tried.
这是我试过的代码。
class Main
{
static class Main2
{
public static void main ( String [ ] args )
{
System . out . println ( "YES!!!!!!!!!!!" ) ;
}
}
public static void main ( String [ ] args )
{
System . out . println ( Main2 . class . getCanonicalName ( ) ) ;
System . out . println ( Main2 . class . getName ( ) ) ;
}
}
class Main3
{
public static void main ( String [ ] args )
{
System . out . println ( "YES!!!!!!!!!!!" ) ;
}
}
- getCanonicalNameand getNamereturn different names. Which one is right? They are both wrong.
- Main3 works.
- getCanonicalName和getName返回不同的名称。哪一个是对的？他们都错了。
- Main3 作品。
回答by Carl Smotricz
I think I see a fix for part of the problem: process.waitFor()prevents control from returning to main() before the subprocess ends.
我想我看到了部分问题的修复：process.waitFor()防止在子进程结束之前控制返回到 main()。
To figure out why your spawned process isn't starting, I'd recommend printing out all the arguments to the ProcessBuilderconstructor and checking that a hand-called JVM called with those arguments succeeds. In particular, you need that class name to be the name of a class having a static void main(String[]).
要弄清楚为什么您的衍生进程没有启动，我建议将所有参数打印到ProcessBuilder构造函数并检查使用这些参数调用的手动调用的 JVM 是否成功。特别是，您需要该类名是具有static void main(String[]).
