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V24.1

XtraReport.ExportToImage(Stream, DXImageFormat) Method

Exports a report to the specified stream in the specified image format.

Namespace: DevExpress.XtraReports.UI

Assembly: DevExpress.XtraReports.v24.1.dll

NuGet Package: DevExpress.Reporting.Core

Declaration

public void ExportToImage(
    Stream stream,
    DXImageFormat format
)

Parameters

Name Type Description
stream Stream

A Stream for output data.

format DXImageFormat

The image format.

Remarks

Note

Once the document export has started, it runs to completion and you cannot interrupt or cancel it.

This method exports a report to an image in the specified image format.

The current report export options are in effect. To get access to the current report export options, use the report’s XtraReport.ExportOptions property. The property contains the ExportOptions object whose ExportOptions.Image property provides access to the ImageExportOptions object that contains the Image export options.

Use the ExportToImageAsync(Stream, ImageExportOptions, CancellationToken) method instead of ExportToImage to export a report asynchronously in a separate task.

Example

This example demonstrates how to export a report to image format.

The project uses the XtraReport.ExportToImage method with the ImageExportOptions object as a parameter.

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

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

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

    // Get its Image export options.
    ImageExportOptions imageOptions = report.ExportOptions.Image;

    // Set Image-specific export options.
    imageOptions.Format = DXImageFormat.Png;

    // Export the report to Image.
    report.ExportToImage(reportPath);

    // Show the result.
    StartProcess(reportPath);
}

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