使用SQL Filestream时出现OutOfMemoryException

时间:2010-12-14 16:03:44

标签: sql filestream

我正在尝试将大约600 MB的zip文件上传到SQL 2008 FILESTREAM表,我得到了OutOfMemoryException。我正在使用SqlFileStream类上传文件(如本教程中所述 - http://www.aghausman.net/dotnet/saving-and-retrieving-file-using-filestream-sql-server-2008.html)。我有一台带有4GB内存的32位Vista机器,如果这很重要,我正在使用VS 2010,Entity Framework 4。

这是我的代码片段 -

public static void AddItem(RepositoryFile repository)
    {
        var contents = repository.Data; // I get the exception at this line.
        repository.Data = System.Text.Encoding.ASCII.GetBytes("0x00");

        using (var scope = new TransactionScope())
        {
            using (var db = new MyEntities())
            {
                db.RepositoryTable.AddObject(repository);
                db.SaveChanges();
            }

            using (var con = new SqlConnection(ConfigurationManager.ConnectionStrings["ConnectionString"].ConnectionString))
            using (var cmd = new SqlCommand("SELECT Data.PathName(), GET_FILESTREAM_TRANSACTION_CONTEXT() FROM dbo.RepositoryTable", con))
            {
                cmd.Connection.Open();
                using (var reader = cmd.ExecuteReader())
                {
                    while (reader.Read())
                    {
                        var path = reader.GetString(0);
                        var transactionContext = reader.GetSqlBytes(1).Buffer;
                        var fileStream = new SqlFileStream(path, transactionContext, FileAccess.Write);

                        fileStream.Write(contents, 0, contents.Length);
                        fileStream.Close();
                    }
                }
            }

            scope.Complete();
        }
    }

如何上传文件而不出错?

谢谢!

2 个答案:

答案 0 :(得分:0)

了解错误发生的确切位置可以帮助找到解决方案。一个可能有帮助的改变是将文件写入块(不要一次写入,循环并一次写入一点)。这使流有机会刷新,从而释放系统资源。

答案 1 :(得分:0)

我想出了这个问题。这是我的代码 -

private void AddFile()
    {
        if (!fupFile.HasFile)
        {
            lblMessage.Text = "Please select a file.";                
            return;
        }

        var data = new byte[(int) fupFile.FileContent.Length];
        fupFile.FileContent.Read(data, 0, data.Length);

        if (fupFile.FileContent.Length > 0)
        {
            var repositoryFile = new Repository
                                     {
                                         ID = Guid.NewGuid(),
                                         Name = Path.GetFileName(fupFile.PostedFile.FileName),                                             
                                         Data = System.Text.Encoding.ASCII.GetBytes("0x00")
                                     };

            RepositoryController.AddItem(repositoryFile, data); // Calling DAL class.
        }
    }

// DAL method
public static void AddItem(RepositoryFile repository, byte[] data)
{
    using (var scope = new TransactionScope())
    {
        using (var db = new MyEntities()) // DBContext
        {
            db.RepositoryTable.AddObject(repository);
            db.SaveChanges();
        }

        using (var con = new SqlConnection(ConfigurationManager.ConnectionStrings["ConnectionString"].ConnectionString))
        using (var cmd = new SqlCommand(string.Format("SELECT Data.PathName(), GET_FILESTREAM_TRANSACTION_CONTEXT() FROM dbo.RepositoryTable WHERE ID='{0}'", repository.ID), con)) // "Data" is the column name which has the FILESTREAM. Data.PathName() gives me the local path to the file.
        {
            cmd.Connection.Open();
            using (var reader = cmd.ExecuteReader())
            {
                while (reader.Read())
                {
                    var path = reader.GetString(0);
                    var transactionContext = reader.GetSqlBytes(1).Buffer;
                    var fileStream = new SqlFileStream(path, transactionContext, FileAccess.Write);

                    fileStream.Write(contents, 0, contents.Length); //Write contents to the file.
                    fileStream.Close();
                }
            }
        }

        scope.Complete();
    }
}