how to get latest uploaded File after uploading a file into database : The Official Microsoft ASP.NET Forums

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nrk_hi

Member

136 Points

how to get latest uploaded File after uploading a file into database

Dec 31, 2012 04:31 AM|LINK

Hi All,
I'm uploading a File into sqlserver using fileupload in file upload Method i m calling a method which gives latest file uploads, but it is not showning the file last file uploaded
after uploading the file then i need to refresh the page then only it is showing the latest file which i uploaded. without refreshing the page how to get the latest file that uploaded in DB From FIleUpload() method. i m writing the code as below :-

protected void Page_Load(object sender, EventArgs e)
    {
        if (Request.QueryString["preview"] == "1" && !string.IsNullOrEmpty(Request.QueryString["fileId"]))
        {
            var fileId = Request.QueryString["fileId"];
            var fileContentType = (string)Session["fileContentType_" + fileId];
            var fileName = (string)Session["fileName_" + fileId];


            string ct = (string)Session["fileContentType_" + fileId];
            using (SqlConnection _con = new SqlConnection("data source=test;Initial Catalog=test;User Id=sa;Password=sql;"))
            using (SqlCommand _cmd = new SqlCommand("UploadFile", _con))
            {
                _cmd.CommandType = CommandType.StoredProcedure;
                _cmd.Parameters.AddWithValue("@FileName", fileName);
                _cmd.Parameters.AddWithValue("@FileType", fileContentType);
                _cmd.Parameters.AddWithValue("@FileContent", (byte[])Session["fileContents_" + fileId]);

                _con.Open();
                _cmd.ExecuteNonQuery();
                _con.Close();
            }

            Response.Clear();
            Response.ContentType = fileContentType;
          
        }

        if (!IsPostBack)
        {
            DataTable fileList = GetFileList();
            gvFiles.DataSource = fileList;
            gvFiles.DataBind();
        }

    }

AND THE GetFileList() mETHOD WHICH GIVES THE UPLOADED FILES :-

public static DataTable GetFileList()
    {
        DataTable fileList = new DataTable();
        using (SqlConnection _con = new SqlConnection("data source=test;Initial Catalog=test;User Id=sa;Password=sql;"))
        {
            _con.Open();
            SqlCommand cmd = new SqlCommand();
            cmd.Connection = _con;
            cmd.CommandTimeout = 0;

            cmd.CommandText = "SELECT ID, FileName FROM Files";
            cmd.CommandType = CommandType.Text;
            SqlDataAdapter adapter = new SqlDataAdapter();

            adapter.SelectCommand = cmd;
            adapter.Fill(fileList);

            _con.Close();
        }

        return fileList;
    }

Ravikumar

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kedarrkulkar...

All-Star

34013 Points

5468 Posts

Re: how to get latest uploaded File after uploading a file into database

Dec 31, 2012 05:10 AM|LINK

because, u r running the display file code inside !IsPostBack condition

remove this

if (!IsPostBack)
        {
            DataTable fileList = GetFileList();
gvFiles.DataSource = fileList;
gvFiles.DataBind();
        }

hope this helps...

Cheers!
KK

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nrk_hi

Member

136 Points

110 Posts

Re: how to get latest uploaded File after uploading a file into database

Dec 31, 2012 05:20 AM|LINK

I Removed ispostback , code is below :-

protected void Page_Load(object sender, EventArgs e)
    {
        if (Request.QueryString["preview"] == "1" && !string.IsNullOrEmpty(Request.QueryString["fileId"]))
        {
            var fileId = Request.QueryString["fileId"];
            var fileContentType = (string)Session["fileContentType_" + fileId];
            var fileName = (string)Session["fileName_" + fileId];


            string ct = (string)Session["fileContentType_" + fileId];
            using (SqlConnection _con = new SqlConnection("data source=test;Initial Catalog=test;User Id=sa;Password=sql;"))
            using (SqlCommand _cmd = new SqlCommand("UploadFile", _con))
            {
                _cmd.CommandType = CommandType.StoredProcedure;
                _cmd.Parameters.AddWithValue("@FileName", fileName);
                _cmd.Parameters.AddWithValue("@FileType", fileContentType);
                _cmd.Parameters.AddWithValue("@FileContent", (byte[])Session["fileContents_" + fileId]);

                _con.Open();
                _cmd.ExecuteNonQuery();
                _con.Close();
            }

            Response.Clear();
            Response.ContentType = fileContentType;
          
        }

        //if (!IsPostBack)
       // {
            DataTable fileList = GetFileList();
            gvFiles.DataSource = fileList;
            gvFiles.DataBind();
      //  }

    }

Ravikumar