I've seen many people saying to use an iframe or a new page to display a ReportViewer control. Is there a way to display the control inline with the rest of my page without using an iframe?
Right-click in the Store Index action method and select Add View as before, select Genre as the Model class, and press the Add button. This tells the Razor view engine that it will be working with a model object that can hold several Genre objects.
You can use .ascx
user controls as partial views with Razor if they inherit from System.Web.Mvc.ViewUserControl
.
In this instance, you can create an ASCX
that contains your ReportViewer
control and the requisite ScriptManager
in your View\Controller
folder:
<%@ Control Language="C#" AutoEventWireup="true" CodeBehind="ReportViewerControl.ascx.cs" Inherits="MyApp.Views.Reports.ReportViewerControl" %>
<%@ Register TagPrefix="rsweb" Namespace="Microsoft.Reporting.WebForms" Assembly="Microsoft.ReportViewer.WebForms, Version=10.0.0.0, Culture=neutral, PublicKeyToken=b03f5f7f11d50a3a" %>
<form id="form1" runat="server">
<div>
<asp:ScriptManager ID="scriptManager" runat="server" EnablePartialRendering="false" />
<rsweb:ReportViewer Width="100%" Height="100%" ID="reportViewer" runat="server" AsyncRendering="false" ProcessingMode="Remote">
<ServerReport />
</rsweb:ReportViewer>
</div>
</form>
In the code-behind, make sure to include the following in the Page_Init
; otherwise, you won't be able to use any options in the report view:
protected void Page_Init(object sender, EventArgs e)
{
// Required for report events to be handled properly.
Context.Handler = Page;
}
You also want to make sure that your control inherits from System.Web.Mvc.ViewUserControl
:
public partial class ReportViewerControl : ViewUserControl
To use this control, you would do something like this in your Razor page:
@Html.Partial("ReportViewerControl", Model)
You can then setup your ReportViewer in the Page_Load
of the control as you normally would. You will have access to an object
named Model
, which you can cast to the type of the model that you send in and then use:
ReportViewParameters model = (ReportViewParameters)Model;
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