运行时问题

Question

如何使这个工作在Windows上，文件filename.txt没有被创建。

Process p = Runtime.getRuntime().exec("cmd echo name > filename.txt"); 

显然，期望输出是一个 “FILENAME.TXT” 应被创建（C：\ Documents和Settings \用户名\ FILENAME.TXT）与内容 “名称”。

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之所以能够用下面的代码来管理，即使该文件是 “FILENAME.TXT” 不被用的ProcessBuilder

 Runtime runtime = Runtime.getRuntime(); 
     Process process = runtime.exec("cmd /c cleartool lsview"); 
     // Directly to file 

//Process p = Runtime.getRuntime().exec( 
//    new String[] { "cmd", "/c", "cleartool lsview > filename.txt" },null, new File("C:/Documents and Settings/username/")); 

     InputStream is = process.getInputStream(); 
     InputStreamReader isr = new InputStreamReader(is); 
     BufferedReader br = new BufferedReader(isr); 
     String line; 

     System.out.printf("Output of running %s is:", 
      Arrays.toString(args)); 

     while ((line = br.readLine()) != null) { 
     System.out.println(line); 
     } 

或使用ProceessBuilder创建，

Process process = new ProcessBuilder("cmd", "/c", "cleartool lsview").start(); 
InputStream is = process.getInputStream(); 
BufferedReader br = new BufferedReader(new InputStreamReader(is)); 

System.out.printf("Output of running %s is:", Arrays.toString(args)); 

String line; 
while ((line = br.readLine()) != null) { 
    System.out.println(line); 
} 

Answer 1

你实际上应使用ProcessBuilder而不是Runtime.exec（请参阅the docs）。

ProcessBuilder pb = new ProcessBuilder("your_command", "arg1", "arg2"); 
pb.directory(new File("C:/Documents and Settings/username/")); 

OutputStream out = new FileOutputStream("filename.txt"); 
InputStream in = pb.start().getInputStream(); 

byte[] buf = new byte[1024]; 
int len; 
while ((len = in.read(buf)) > 0) 
    out.write(buf, 0, len); 

out.close(); 

（我会适应它cmd并呼应，如果我有一个窗口机在到达...随意编辑这个职位！）

Answer 2

它应该与

Process p = Runtime.getRuntime().exec(
    new String[] { "cmd", "/c", "echo name > filename.txt" }); 

我目前没有运行Windows，所以很不幸我无法测试它。

背后的原因是，在您的版本中，命令在每个空格字符处被分割。因此，运行时所要做的是创建一个进程cmd并为其提供参数echo,name,>和filename.txt，这是没有意义的。命令echo name > filename.txt是cmd进程的单个参数，因此您必须手动为不同参数提供一个数组。

如果你想确保该文件是在一个特定的文件夹中创建你必须提供一个工作目录exec()仅在三个参数版本的工作原理：

Process p = Runtime.getRuntime().exec(
    new String[] { "cmd", "/c", "echo name > filename.txt" }, 
    null, new File("C:/Documents and Settings/username/"));