Download multiple Xml file

[Tash18]

Hi guyzz.. Im developing a web application here i get datas from the user and on a button click these datas get stored in my data base now i have another button on clicking it,it retrieves all the data from the database and gets saved in my system as an XML file... the code is as follows for the button that creates an xml file...

private void Button1_Click(object sender, System.EventArgs e)
{
SqlDataAdapter da = null;
try
{
con = new SqlConnection();
con.ConnectionString = ConfigurationSettings.AppSettings["DSN"];
da = new SqlDataAdapter("select * from card_details",con);
DataSet ds = new DataSet();
da.Fill(ds,"card_details");
ds.WriteXml("D:\ADIB\OutputXML.xml");
}
catch(SqlException ex)
{
throw new System.Exception ("Error -- " +ex.Message);
}
catch(Exception exp)
{
throw new System.Exception ("Error -- " +exp.Message);
}
finally
{
con.Close();
da.Dispose();
}
}

Now my issue is i need my application to create the xml file,before saving it into my system it should ask me whether to open the file or save the file so that i can specify the file name (something similar to a download window in our web browser)... please advice me... Thanx in advance.. Regards, Tash

[Not Active]

Maybe use something like SaveFileDialog[^] :rolleyes:

---

I know the language. I've read a book. - _Madmatt

[Tash18]

Im using VS2003 framework1.1 Any other ideas or any other ways do please let me know.... Thanx in advance Regards, Tash

modified on Wednesday, February 3, 2010 11:05 PM

[Tash18]

Any ideas or any ways do please let me know...... Regards, Tash

[Tash18]

Sorry it dint work.. Im using VS2003... plz advice me... Thanx in advance.. Regards, Tash

[Palash Biswas]

use below code

private void button1_Click(object sender, EventArgs e)
{
saveFileDialog1.FileOk += new CancelEventHandler(saveFileDialog1_FileOk);
saveFileDialog1.ShowDialog(this);

    }

    void saveFileDialog1\_FileOk(object sender, CancelEventArgs e)
    {
        SaveFileDialog ss = sender as SaveFileDialog;

        SqlDataAdapter da = null;
        try
        {
            con = new SqlConnection();
            con.ConnectionString = ConfigurationSettings.AppSettings\["DSN"\];
            da = new SqlDataAdapter("select \* from card\_details", con);
            DataSet ds = new DataSet(); da.Fill(ds, "card\_details");
            ds.WriteXml(ss.FileName);
        }
        catch (SqlException ex)
        {
            throw new System.Exception("Error -- " + ex.Message);
        }
        catch (Exception exp)
        {
            throw new System.Exception("Error -- " + exp.Message);
        }
        finally
        {
            con.Close(); da.Dispose();
        }
        
    }

saveFileDialog1 is SaveFileDialog type win control.

[Tash18]

Hi thanx for ur reply... My problem is not solved yet... as u said when i type saveFileDialog1. i dont find "FileOk"... i dont know why... may be its because im using VS2003??? plz help me... Thanx in advance... Regards,\ Tash

[Ravindra Nidhonkar]

Following code should solve your problem:

Response.Clear();
Response.ContentType = "text/plain";
Response.WriteFile("D:\\ADIB\\OutputXML.xml");
Response.Flush();

[Palash Biswas]

I wrote my code in VS 2008. Under dialog section of toolbox you can get the specified control. TRY this Code

System.Windows.Forms.SaveFileDialog saveFileDialog = new System.Windows.Forms.SaveFileDialog();

saveFileDialog.Filter = "XML Files (*.xml)|*.*";

if (saveFileDialog.ShowDialog(this) == DialogResult.OK)

{

string FileName = saveFileDialog.FileName;

if (FileName.Trim() != "")

{

// your Code here
}

}

[Tash18]

I did as u said.. but got an error in this line --->

if (saveFileDialog.ShowDialog(this) == DialogResult.OK)

The error stated was ---> The best overloaded method match for 'System.Windows.Forms.CommonDialog.ShowDialog(System.Windows.Forms.IWin32Window)' has some invalid arguments Thanx in advance... Regards, Tash

[Palash Biswas]

try this

if (saveFileDialog.ShowDialog() == DialogResult.OK)

[Tash18]

Hi.. now ther was no build errors but after execution.. when i click my button to create Xml file i receive the following exception ---> It is invalid to show a modal dialog or form when the application is not running in UserInteractive mode. Specify the ServiceNotification or DefaultDesktopOnly style to display a notification from a service application. And the line in which this exception ocuurs -->>

if (saveFileDialog.ShowDialog() == DialogResult.OK)

thanx in advance.. Regards, Tash

[Palash Biswas]

I think you are not use win form application . Is that true if it is then you can't do it using previous procedure.Please use a Custom pop up to do this.

[Tash18]

Ohh no.. so what am i supposed 2 do??? any idea??? regards, tash

[Palash Biswas]

This will help you much , I think it works fine :-D

Response.ContentType = "text/xml";
Response.AppendHeader("Content-Disposition", "attachment; filename=mmm.xml");
Response.TransmitFile(Server.MapPath("~/images/mmm.jpg"));
SqlDataAdapter da = null;
try
{
con = new SqlConnection();
con.ConnectionString = ConfigurationSettings.AppSettings["DSN"];
da = new SqlDataAdapter("select * from card_details", con);
DataSet ds = new DataSet(); da.Fill(ds, "card_details");
ds.WriteXml(Response.OutputStream);
Response.End();
}
catch (SqlException ex)
{
throw new System.Exception("Error -- " + ex.Message);
}
catch (Exception exp)
{
throw new System.Exception("Error -- " + exp.Message);
}
finally
{
con.Close(); da.Dispose();
}

modified on Thursday, February 4, 2010 6:01 AM

[Palash Biswas]

Try this i think it will help you; Response.ContentType = "text/xml"; Response.AppendHeader("Content-Disposition", "attachment; filename=mmm.xml"); Response.TransmitFile(Server.MapPath("~/images/sailbig.jpg")); SqlDataAdapter da = null; try { con = new SqlConnection(); con.ConnectionString = ConfigurationSettings.AppSettings["DSN"]; da = new SqlDataAdapter("select * from card_details", con); DataSet ds = new DataSet(); da.Fill(ds, "card_details"); ds.WriteXml(Response.OutputStream); Response.End(); } catch (SqlException ex) { throw new System.Exception("Error -- " + ex.Message); } catch (Exception exp) { throw new System.Exception("Error -- " + exp.Message); } finally { con.Close(); da.Dispose(); } }

[Palash Biswas]

I think , The below code will help you much

Response.ContentType = "text/xml";
Response.AppendHeader("Content-Disposition", "attachment; filename=mmm.xml");
Response.TransmitFile(Server.MapPath("~/images/mmm.jpg"));
SqlDataAdapter da = null;
try
{
con = new SqlConnection();
con.ConnectionString = ConfigurationSettings.AppSettings["DSN"];
da = new SqlDataAdapter("select * from card_details", con);
DataSet ds = new DataSet(); da.Fill(ds, "card_details");
ds.WriteXml(Response.OutputStream);
Response.End();
}
catch (SqlException ex)
{
throw new System.Exception("Error -- " + ex.Message);
}
catch (Exception exp)
{
throw new System.Exception("Error -- " + exp.Message);
}
finally
{
con.Close(); da.Dispose();
}

[Palash Biswas]

use ...

Response.ContentType = "text/xml";
Response.AppendHeader("Content-Disposition", "attachment; filename=mmm.xml");
Response.TransmitFile(Server.MapPath("~/images/mmm.jpg"));
SqlDataAdapter da = null;
try
{
con = new SqlConnection();
con.ConnectionString = ConfigurationSettings.AppSettings["DSN"];
da = new SqlDataAdapter("select * from card_details", con);
DataSet ds = new DataSet(); da.Fill(ds, "card_details");
ds.WriteXml(Response.OutputStream);
Response.End();
}
catch (SqlException ex)
{
throw new System.Exception("Error -- " + ex.Message);
}
catch (Exception exp)
{
throw new System.Exception("Error -- " + exp.Message);
}
finally
{
con.Close(); da.Dispose();
}