android從assets文件夾復制數據庫

Question

我無法將數據庫從資產文件夾復制到數據庫文件夾。 當用戶啟動應用程序時，我檢查數據庫是否不存在，如果為true，則復制數據庫。

我的代碼在哪里：

    private void CopyDatabaseIfNotExists() {

        dbName = "quizdb.db";

        File f = getDatabasePath(dbName);
        if (f.exists())
            return;

        System.out.println("db missing");

        try {

            InputStream mInputStream = getAssets().open(dbName);

            OutputStream mOutputStream = new FileOutputStream(f);
            byte[] buffer = new byte[1024];
            int length;
            while ((length = mInputStream.read(buffer)) > 0) {
                mOutputStream.write(buffer, 0, length);
            }
            mOutputStream.flush();
            mOutputStream.close();
            mInputStream.close();

        } catch (NullPointerException e) {
            e.printStackTrace();
        } catch (FileNotFoundException e) {
            e.printStackTrace();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

我得到了這個錯誤：

06-15 12:29:04.882  25037-25037/com.ex.example W/System.err﹕ java.io.FileNotFoundException: /data/data/com.ex.example/databases/quizdb.db: open failed: ENOENT (No such file or directory)

我已經嘗試搜索解決方案，但找不到。 有人可以幫助我嗎？ 謝謝，對不起我的英語。

Answer 1

試試這個...對我有用.. !! 並且您確定已將數據庫文件放入資產文件夾中。

 private void copyDataBase() throws IOException {
    InputStream is = myContext.getAssets().open(DB_NAME);
    // Log.v("Tag assets",is.toString());
    String outFileName = DB_PATH + DB_NAME;
    OutputStream out = new FileOutputStream(outFileName);
    Log.v("Tag assets", out.toString());
    byte[] buffer = new byte[1024];
    int length;
    while ((length = is.read(buffer)) > 0) {
        // Log.v("Tag",out.toString());
        out.write(buffer, 0, length);
        // Log.v("Tag",out.toString());
    }
    // Log.v("Tag","Database created");
    is.close();
    out.flush();
    out.close();      

}

Answer 2

當文件不存在且無法創建時， FileOutputStream引發此異常。 我之前有這個探針，並設法通過在OutputStream mOutputStream = new FileOutputStream(f);之前先調用Context對象（或SQLiteDatabase類）的openOrCreateDatabase方法來解決它OutputStream mOutputStream = new FileOutputStream(f);