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运行时问题

runtime.exec java

2

srinannapa · 技术社区 · 14 年前

如何在windows上工作,文件文件名.txt未创建。

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

很明显,预期的产出是文件名.txt“应创建(C:\Documents and Settings\用户名\文件名.txt)内容为“名称”。

即使文件文件名.txt“”不是用processBuilder创建的

       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);
       }

或者,使用ProceessBuilder,

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);
}

2 回复 | 直到 12 年前

1

6

aioobe 14 年前

ProcessBuilder 而不是 Runtime.exec (见 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();

(如果我能接触到一台windows机器的话,我会把它改编成cmd和echo。。。请随意编辑此帖子!)

2

4

musiKk 14 年前

它应该与

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

我现在没有运行Windows,所以很不幸我不能测试它。

这背后的原因是,在您的版本中,命令在每个空格字符处被拆分。所以运行时所做的就是创建一个进程 cmd 并为它提供论据 echo , name , > 和 filename.txt 这毫无意义。命令 echo name > filename.txt 是一个单一的论点

exec() 仅适用于三参数版本:

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