从iPad应用程序中的Sqlite数据库中检索BLOB图像时出现问题

Question

我正在编写一段代码来存储和检索照片作为BLOB到我的sqlite数据库。

照片正在成功添加到数据库中，但是当我检索图像时，它不会显示在UIImageView中。

有人能看到这里有什么问题吗？

这是存储图像的代码：

 if (sqlite3_open([[self filePath] UTF8String], &db) == SQLITE_OK) {    

     NSData *imgData = UIImagePNGRepresentation(TempImg); //TempImg is an UIImage instance
     char *err;
     NSString *sql = [NSString stringWithFormat:@"INSERT INTO 'Picture' ('Pic', 'PID') VALUES ('%@', '%@')",imgData, PID];
     sqlite3_stmt *statement;

    sqlite3_bind_blob(sql, 0, [imgData bytes], [imgData length], nil);  

     if (sqlite3_prepare_v2( db, [sql UTF8String], -1, &statement, nil) == SQLITE_OK) {

         if (sqlite3_exec(db, [sql UTF8String], NULL, NULL, &err)!= SQLITE_OK )
         {sqlite3_close(db);}

         else{
             UIAlertView *alert = [[UIAlertView alloc] 
                                   initWithTitle:@"Success" message:@"The Photo have been successfully added." delegate:self cancelButtonTitle:@"OK" otherButtonTitles:nil];
             [alert show];
             [alert release];

             sqlite3_close(db);
         }
     }
 }

这是检索代码：

    if (sqlite3_open([[self filePath] UTF8String], &db) == SQLITE_OK) {         
        NSString *sql = [NSString stringWithFormat:@"SELECT * FROM Picture Where PicID = '14'"];
        sqlite3_stmt *statement;

        if (sqlite3_prepare_v2( db, [sql UTF8String], -1, &statement, nil) == SQLITE_OK) {
            while (sqlite3_step(statement) == SQLITE_ROW)
            {
                char *field1 = (char *) sqlite3_column_text(statement, 0);//the name of the picture
                NSString *field1Str = [[NSString alloc] initWithUTF8String: field1];
                field.text= field1Str;

                NSData *data = [[NSData alloc] initWithBytes:sqlite3_column_blob(statement, 1) length:sqlite3_column_bytes(statement, 1)];

if (data==nil) {
    UIAlertView *alert = [[UIAlertView alloc] 
                          initWithTitle:@"NULL" message:@"" delegate:self cancelButtonTitle:@"OK" otherButtonTitles:nil];
    [alert show];
    [alert release];
}
                UIImage *newImage = [UIImage imageWithData:data]; 
                [Img setImage:newImage];//Img is a UIImageView instance             
            }
            sqlite3_close(db);
                 }
}

问题是，Img UIImageView没有显示任何内容。