在android中创建临时文件

Question

在我的android应用程序中，我试图用create temp创建一个音乐文件，并尝试读取该文件。现在我能够创建，但它给出了一个错误，指出无法播放视频。

public void onCreate(Bundle savedInstanceState) { 

    super.onCreate(savedInstanceState); 
    requestWindowFeature(Window.FEATURE_NO_TITLE); 
    //File newfile=new File("/data/data/com.ayansys.samplevideo/Music.3gp"); 

    //String xyz=newfile.getAbsolutePath(); 
    getWindow().setFlags(WindowManager.LayoutParams.FLAG_FULLSCREEN,WindowManager.LayoutParams.FLAG_FULLSCREEN); // Sets the window in full screen 
    setContentView(R.layout.main); 
    // path.g 
    // Log.i("Raw folder path:", R.raw.music); 
    String path= musicFile(); 
    try{ 
    Thread.sleep(5000); 
    } 
    catch(InterruptedException e) 
    { 
     e.printStackTrace(); 
     Log.i("exception raised",e.toString()); 
    } 
    mVideoView = (VideoView) findViewById(R.id.surface_view); 
    // mVideoView.setVideoURI(Uri.parse("android.resource://com.ayansys.samplevideo/"+R.raw.music)); 

    // File infile=new File("/data/data/com.ayansys.samplevideo/Video.3gp"); 
    mVideoView.setVideoPath(path); 

    mVideoView.requestFocus(); 
    mVideoView.start(); 

} 

public String musicFile() 
{ 
    String temPath = null; 
    try{ 
     File outFile=File.createTempFile("music", ".3gp", getDir(temPath, MODE_PRIVATE)); 
     temPath=outFile.getAbsolutePath(); 
     FileOutputStream f=new FileOutputStream(outFile); 
     //FileOutputStream fos = openFileOutput("music.3gp", Context.MODE_PRIVATE); 
     URL u = new URL("http://59.162.166.211/aptv-web/VODAFONE/STARONE/SarabhaivsSarabhai/J2ME/SarabhaivsSarabhai_05.3gp"); 
    HttpURLConnection c = (HttpURLConnection) u.openConnection(); 
    c.setRequestMethod("GET"); 
    c.setDoOutput(true); 
    c.connect(); 
    //FileOutputStream f = new FileOutputStream(new File("/sdcard/videooutput.mp4")); 


    InputStream in = c.getInputStream(); 

    byte[] buffer = new byte[1024]; 
    int len1 = 0; 
    while ((len1 = in.read(buffer))!= -1) { 
     f.write(buffer,0,len1); 
    } 

    f.close(); 
    String str=outFile.getAbsolutePath(); 
    //outFile..close(); 
    in.close(); 

} 
    catch(Exception e) 
    { 
     e.printStackTrace(); 
     System.out.println("Exception raised in creating a file: "+e.toString()); 
    } 
    return temPath; 
} 

}

我想玩这个。 请分享您宝贵的建议。提前

Answer 1

Context.openFileOutput(String name, int mode)打开的内部存储文件

感谢。

我建议您阅读Android Data Storage Guide，因为它涵盖了内部和外部存储问题。