Resources.openRawResource（）问题的Andr​​oid [英] Resources.openRawResource() issue Android

问题描述

我在 RES /生/ 文件夹中的数据库文件。我打电话 Resources.openRawResource（）的文件名如 R.raw.FileName ，我得到的输入流，但我有在设备中的另一个数据库文件，所以数据库我用的数据库中的内容复制到设备：

I have a database file in res/raw/ folder. I am calling Resources.openRawResource() with the file name as R.raw.FileName and I get an input stream, but I have an another database file in device, so to copy the contents of that db to the device db I use:

 BufferedInputStream bi = new BufferedInputStream(is);

和FileOutputStream中，但我得到一个异常的数据库文件已损坏。我该如何进行？ 我尝试使用文件和的FileInputStream 和路径为 / RES /生读取文件/文件名，但也不能正常工作。

and FileOutputStream, but I get an exception that database file is corrupted. How can I proceed? I try to read the file using File and FileInputStream and the path as /res/raw/fileName, but that also doesn't work.

推荐答案

是的，你应该能够使用 openRawResource 从您的原始资源文件夹复制跨二进制设备

Yes, you should be able to use openRawResource to copy a binary across from your raw resource folder to the device.

InputStream ins = getResources().openRawResource(R.raw.my_db_file);
ByteArrayOutputStream outputStream=new ByteArrayOutputStream();
int size = 0;
// Read the entire resource into a local byte buffer.
byte[] buffer = new byte[1024];
while((size=ins.read(buffer,0,1024))>=0){
  outputStream.write(buffer,0,size);
}
ins.close();
buffer=outputStream.toByteArray();

你的文件的副本现在应该在缓冲存，这样你就可以使用的FileOutputStream 保存缓冲区到一个新的文件中。

A copy of your file should now exist in buffer, so you can use a FileOutputStream to save the buffer to a new file.

FileOutputStream fos = new FileOutputStream("mycopy.db");
fos.write(buffer);
fos.close();

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