Android - 从资产中复制数据库

时间:2013-07-10 12:06:00

标签: android database sqlite fileoutputstream

以下是我正在使用的代码(在许多答案中找到):

InputStream myInput;
        try {
            myInput = iNezamApplication.getAppContext().getAssets().open(DB_NAME);
            String outFileName = DB_PATH + DB_NAME;

            OutputStream myOutput = new FileOutputStream(outFileName);
            byte[] buffer = new byte[1024];
            int length;
            while ((length = myInput.read(buffer))>0){
                myOutput.write(buffer, 0, length);
            }

            myOutput.flush();
            myOutput.close();
            myInput.close();

        } catch (IOException e1) {
            e1.printStackTrace();

        }

但是,在到达OutpoutStream行后我总是遇到异常:

java.io.FileNotFoundException: /data/data/package_name/databases/databasename.db: open failed: ENOENT (No such file or directory)

3 个答案:

答案 0 :(得分:6)

我试过这样的事情......

final String[] sample_dbName = {"DrugsNew.db"};

    int assetDbSize = sample_dbName.length;
    File databaseFile = new File( "/data/data/com.handyrep2.ui/databases");

    // check if databases folder exists, if not create one and its subfolders
    if (!databaseFile.exists()){
        databaseFile.mkdir();
    }

        for(int i=0;i<assetDbSize;i++){
            String outFilename =null;

        outFilename = "/data/data/com.handyrep2.ui/databases"+ "/" + sample_dbName[i];

            File sampleFile = new File(outFilename);

                try {
                    InputStream in = activity.getAssets().open("offlinedb/"+sample_dbName[i]);

                    OutputStream out = new FileOutputStream(outFilename);
                    // Transfer bytes from the sample input file to the sample output file
                    byte[] buf = new byte[1024];
                    int len;
                    while ((len = in.read(buf)) > 0) {
                      out.write(buf, 0, len);
                    }
                    out.flush();
                    // Close the streams
                    out.close();
                    in.close();
                }catch(IOException e) {

                }


        }

答案 1 :(得分:1)

“package_name”是否可以替换您的真实包名,只是在这里发布或在DB_PATH中真正使用它? :)

答案 2 :(得分:1)

您必须先创建一个File对象