Question

好的，我们走了：当我通过static main传递参数时，我的程序运行正常。但是，当我传递相同的“论点”在主要内部声明时，不起作用。如果声音混乱，这是代码。我想要的只是做这些事情，没有args通过main。这是DEBUG：http://i.stack.imgur.com/sbGyo.png

import java.io.*;
import java.util.*;

public class DoProcessBuilder extends Thread {
  public static void main (String[] args) throws IOException, InterruptedException {
    DoProcessBuilder teste = new DoProcessBuilder();

    String[] uia = {"ls","-al","|","grep","bash"};
    teste.ExecCommand(uia);    // here this not works, WHY? if I execute the "java DoProcessBuilder ls -al | grep bash" works fine?

    teste.ExecCommand(args); // works fine!

  }

  public  String ExecCommand(String args[]) throws IOException, InterruptedException {
    StringBuffer x = new StringBuffer();
    if (args.length <= 0) {
        System.err.println("Need command to run");
    }

    Process process = new ProcessBuilder(args).start();
    process.waitFor();
    InputStream is = process.getInputStream();
    InputStreamReader isr = new InputStreamReader(is);
    BufferedReader br = new BufferedReader(isr);
    String line = "";

    while((line = br.readLine()) != null){
        x.append(line+"\n");
    }
    System.out.println("\nCOMMAND OUT \n"+x.toString());

    return x.toString();

  }
}

Answer 1

您正在尝试使用ProcessBuilder执行shell管道。那不行。管道|不是ls命令的有效参数。您必须启动一个shell new ProcessBUilder("sh")。

它与args一起工作正常，因为你的程序没有得到管道，其余来自调用shell。您的计划只获得ls -al。

Answer 2

没有问题，你提到的地方。但是使用teste.ExecCommand(args);，您将获得ArrayIndexOutOfBoundsException

   StringBuffer x = new StringBuffer();
    if (args.length <= 0) {
        System.err.println("Need command to run");
    }

    Process process = new ProcessBuilder(args).start(); // ArrayIndexOutOfBoundsException
    process.waitFor();
    InputStream is = process.getInputStream();
    InputStreamReader isr = new InputStreamReader(is);
    BufferedReader br = new BufferedReader(isr);
    String line = "";

    while((line = br.readLine()) != null){
        x.append(line+"\n");
    }
    System.out.println("\nCOMMAND OUT \n"+x.toString());

    return x.toString();

main（String [] args）不同的String [] args = {.....}？

2 个答案: