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XtraReport.ExportToRtf(String) Method

Exports a report to the specified file path in RTF format.

Namespace: DevExpress.XtraReports.UI

Assembly: DevExpress.XtraReports.v19.1.dll


public void ExportToRtf(
    string path
Public Sub ExportToRtf(
    path As String


Name Type Description
path String

A String which specifies the file name (including the full path) for the created RTF file.


Once the document export has started, it will run until the resulting document is complete and cannot be interrupted or canceled in the process.

Use this method to export a report to an RTF file using its current RTF-specific export options. These options are represented by the RtfExportOptions object returned by the ExportOptions.Rtf property of a report's XtraReport.ExportOptions.


This method overwrites any files present at the specified path that have the same file name without any warnings.


For more information on using the ExportToRtf method, and limitations on export to RTF, please refer to the Export to RTF document.


The following example demonstrates how to export a report to RTF format. To do this, the XtraReport.ExportToRtf method should be used.

For this example to work, you first need to add a new report (named XtraReport1) to your project, drop some report controls onto it, and then execute the code below (for instance, in the Button.Click event handler of any button on a form). Note also, that if you want the resulting file to be automatically opened in the default program which is used to open *.rtf files in your system, you can call the StartProcess method, which is also shown in this example.

using System.Diagnostics;
using DevExpress.XtraPrinting;
using DevExpress.XtraReports.UI;
// ...

private void ExportToRTF()
    // A path to export a report.
    string reportPath = "c:\\Test.rtf";

    // Create a report instance.
    XtraReport1 report = new XtraReport1();

    // Export the report to RTF.

    // Show the result.

// Use this method if you want to automaically open
// the created RTF file in the default program.
public void StartProcess(string path)
    Process process = new Process();
        process.StartInfo.FileName = path;
    catch { }
See Also