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Resources.openRawResource() issue Android

Tags:

java

android

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);

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.

like image 292
Sam97305421562 Avatar asked Jun 02 '09 12:06

Sam97305421562


1 Answers

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

Based on the example code in the API demos (content/ReadAsset), you should be able to use a variation of the following code snippet to read the db file data.

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();

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();
like image 133
Reto Meier Avatar answered Nov 18 '22 02:11

Reto Meier