如何使用apache commons-exec運行java程序？

Question

I am trying to run java code dynamically in my java application GUI. I have tried the following code: 

      Sring tempfile="java -classpath "+wrkdir+"/bin "+runfile; 
      CommandLine cmdLine = CommandLine.parse(tempfile); 
      DefaultExecuteResultHandler resultHandler = new DefaultExecuteResultHandler(); 
      ExecuteWatchdog watchdog = new ExecuteWatchdog(ExecuteWatchdog.INFINITE_TIMEOUT); 
      DefaultExecutor executor = new DefaultExecutor(); 
      executor.setExitValue(1); 
      executor.setWatchdog(watchdog); 
      try { 
       executor.execute(cmdLine, resultHandler); 
      } catch (ExecuteException e) { 
       // TODO Auto-generated catch block 
       e.printStackTrace(); 
      } catch (IOException e) { 
       // TODO Auto-generated catch block 
       e.printStackTrace(); 
      } 
      try { 
       resultHandler.waitFor(); 
      } catch (InterruptedException e) { 
       // TODO Auto-generated catch block 
       e.printStackTrace(); 
      } 

結果是，當我的輸入文件（tempfile）由打印語句組成時;即，

public class Sample2 { 
    public static void main(String[] args) {        
       System.out.println("It Works..!!"); 
    } 
} 

它能夠顯示結果。但是如果輸入文件是類似的，

import java.io.DataInputStream; 
    import java.io.IOException; 
    import java.util.*; 
    public class Count 
    { 
    public static void main(String args[]) throws IOException 
    { 
     int n; 
     System.out.println("Enter the number: "); 
     DataInputStream din=new DataInputStream(System.in); 
     String s=din.readLine(); 
     n=Integer.parseInt(s); 
     System.out.println("#"+n); 
    } 
    } 

NumberFormatException是結果。這是什麼原因？在這種情況下，我怎樣才能通過鍵盤輸入數值？

Answer 1

您是否認爲輸入文件中的parseInt實際上可能會引發合法的解析異常？

嘗試更改以下行

n=Integer.parseInt(s); 

到

try { 
n=Integer.parseInt(s); 
} catch(NumberFormatException e) { 
    System.out.println("Unable to parse string into Integer. String: " + s); 
} 

Answer 2

的DefaultExector()構造並不重視標準輸入，這樣你就不會得到任何輸入，並沒有什麼解析。要附加到標準輸入，創建一個ExecuteStreamHandler並將其添加到您的DefaultExecutor這樣的：

ExecuteStreamHandler streamHandler = new PumpStreamHandler(System.out, System.err, System.in); 
DefaultExecutor executor = new DefaultExecutor(); 
executor.setStreamHandler(streamHandler); 

如果你想從別人的而不是System.in某處讀取你的輸入，通過適當的InputStream對象的PumpStreamHandler構造。