Java重生过程

发布于 2024-09-05 22:22:11 字数 329 浏览 6 评论 0原文

我正在制作一个类似编辑器的程序。如果用户在主窗口中选择“文件”->“打开”，我想以所选文件名作为参数启动编辑器进程的新副本。但是，为此我需要知道使用什么命令来启动第一个进程：

java -jar myapp.jar blabalsomearguments // --- need this information
> Open File (fileUrl)
> exec("java -jar myapp.jar blabalsomearguments fileUrl");

我不是在寻找进程内解决方案，我已经实现了该解决方案。我希望获得单独流程带来的好处。

原文

I'm making an editor-like program. If the user chooses File->Open in the main window I want to start a new copy of the editor process with the chosen filename as an argument. However, for that I need to know what command was used to start the first process:

java -jar myapp.jar blabalsomearguments // --- need this information
> Open File (fileUrl)
> exec("java -jar myapp.jar blabalsomearguments fileUrl");

I'm not looking for an in-process solution, I've already implemented that. I'd like to have the benefits that seperate processes bring.

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弥繁 2024-09-12 22:22:11

由于您正在启动 Java -> Java，可以使用现有的类路径在命令行上设置类路径。这种事情在开发环境中也非常有效。

ProcessBuilder selfLauncher = new ProcessBuilder(
 "java", "-cp", System.getProperty("java.class.path"),
 "com.my.mainClass" );
selfLauncher.start();

更新：
对于可执行的 jar 文件，您将有一个类路径，它只是 jar 文件本身的相对路径。如果您需要命令行参数，则必须从 main 中保存它们，并在启动时重新应用它们。

您可以通过将以下程序打包到 jar 中来看到这一点。我实际上不确定如果可执行 jar 文件中有 jar 会发生什么。它们可能出现在类路径中。

public class TestJarPath {
    public static void main(String args[]) throws Exception {
        for (String s : args)
         System.out.print("[" + s + "] ");
        System.out.println();

        String cp = System.getProperty("java.class.path");
        for (String s : cp.split(";"))
          System.out.println(s);
    } 
}

对于 java -jar ..\tst.jar X，您会得到如下输出：

[X]
..\tst.jar

Since you are launching Java -> Java, you can use the existing classpath to set the classpath on the command line. This type of thing works really nice in the dev environment too.

ProcessBuilder selfLauncher = new ProcessBuilder(
 "java", "-cp", System.getProperty("java.class.path"),
 "com.my.mainClass" );
selfLauncher.start();

Update:
For executable jar files, you will have a classpath which is simply the relative path to the jar file itself. If you want the command line arguments, you will have to save them from main, and re-apply them when launching.

You can see this by packing the following program into a jar. I'm not actually sure what happens if you have jars inside the executable jar file. They probably show up in the classpath.

public class TestJarPath {
    public static void main(String args[]) throws Exception {
        for (String s : args)
         System.out.print("[" + s + "] ");
        System.out.println();

        String cp = System.getProperty("java.class.path");
        for (String s : cp.split(";"))
          System.out.println(s);
    } 
}

For java -jar ..\tst.jar X, you get output like: