Question

我编写了以下用于将视频字节插入数据库的代码的 DBAdapter.java

public void insertVideoBytesInVideoDownloadsTable(int id, byte[] videoBytes){
    ContentDBHelper dbHelper = new ContentDBHelper(context);
    sqliteDB = dbHelper.getWritableDatabase();
    String sqlInsertVideoBytes = "Insert into " + tbl_video_downloads + " values (?, ?)";
    Log.i(TAG, sqlInsertVideoBytes);
    SQLiteStatement insertStatement = sqliteDB.compileStatement(sqlInsertVideoBytes);
    insertStatement.clearBindings();
    insertStatement.bindLong(1, id);
    Log.i(TAG, "Inserted: "+id);
    insertStatement.bindBlob(2, videoBytes);
    Log.i(TAG, "Inserted: "+videoBytes);
    insertStatement.executeInsert();
    Log.i(TAG, "Execute inserted");
   }

StoreVideoInDB.java

   public String downloadAndStoreVideoInDB(String urlPath){
        try {
            Log.i(TAG , "URL " +urlPath);
            URL url = new URL(urlPath);
            URLConnection connection = url.openConnection();
            connection.connect();
            InputStream videoStream = connection.getInputStream();
            BufferedInputStream videoBufferedStream = new BufferedInputStream(videoStream,128);
            ByteArrayBuffer videoByteArray = new ByteArrayBuffer(500);

            //Get the bytes one by one
            int current = 0;
            while((current = videoBufferedStream.read())!= -1){
                videoByteArray.append((byte)current);
                //Log.i(TAG, String.valueOf(current));
            }
            dbAdapter.insertVideoBytesInVideoDownloadsTable(id, videoByteArray.toByteArray());
        }catch (IOException ex) {
            Log.e(TAG, "error: " + ex.getMessage(), ex);
            ex.printStackTrace();
        }
        return path;
 }

DDMS日志：

01-16 16:03:13.563: INFO/DronaDBAdapter(18235): Insert into video_downloads values (?, ?)
01-16 16:03:13.563: INFO/DBAdapter(18235): Inserted: 1
01-16 16:03:13.573: INFO/DBAdapter(18235): Inserted: [B@44f35120
01-16 16:03:13.653: INFO/DBAdapter(18235): Execute inserted

根据DDMS，值1（ID）和[B @ 44f35120（视频Blob）值显示成功。但它们不会插入数据库中。

当我检查数据库并查询时如下：

select * from video_downloads;

结果：

1 |

第二个字段是空白的！没有插入blob值！为什么？我的代码中有什么问题吗

Answer 1

我认为你误读了select的输出。您的数据存在，但“选择”没有显示，因为它不知道如何处理blob。输出它的字节对于大多数类型的blob来说毫无意义。

尝试从代码中读回数据，它就在那里。

为什么blob值没有存储在android中的sqlite数据库中

1 个答案: