How do I get the HTML output of a UserControl in .NET (C#)?

C#Html.NetUser Controls

C# Problem Overview


If I create a UserControl and add some objects to it, how can I grab the HTML it would render?

ex.

UserControl myControl = new UserControl();
myControl.Controls.Add(new TextBox());

// ...something happens

return strHTMLofControl;

I'd like to just convert a newly built UserControl to a string of HTML.

C# Solutions


Solution 1 - C#

You can render the control using Control.RenderControl(HtmlTextWriter).

Feed StringWriter to the HtmlTextWriter.

Feed StringBuilder to the StringWriter.

Your generated string will be inside the StringBuilder object.

Here's a code example for this solution:

string html = String.Empty;
using (TextWriter myTextWriter = new StringWriter(new StringBuilder()))
{
    using (HtmlTextWriter myWriter = new HtmlTextWriter(myTextWriter))
    {
        myControl.RenderControl(myWriter);
        html = myTextWriter.ToString();
    }
}

Solution 2 - C#

//render control to string
StringBuilder b = new StringBuilder();
HtmlTextWriter h = new HtmlTextWriter(new StringWriter(b));
this.LoadControl("~/path_to_control.ascx").RenderControl(h);
string controlAsString = b.ToString();

Solution 3 - C#

UserControl uc = new UserControl();
MyCustomUserControl mu = (MyCustomUserControl)uc.LoadControl("~/Controls/MyCustomUserControl.ascx");

TextWriter tw = new StringWriter();
HtmlTextWriter hw = new HtmlTextWriter(tw);

mu.RenderControl(hw);
            
return tw.ToString();

Solution 4 - C#

Seven years late, but this deserves to be shared.

The generally accepted solution - StringBuilder into StringWriter into HtmlWriter into RenderControl - is good. But there are some gotchas, which I unfortunately ran across while trying to do the same thing. Some controls will throw errors if they're not inside of a Page, and some will throw errors if they're not inside of a <form> with runat="server". The ScriptManager control exhibits both of these behaviours.

I eventually found a workaround here. The gist of it is basically just instantiating a new Page and Form before doing the writer work:

Page page = new Page();
page.EnableEventValidation = false;

HtmlForm form = new HtmlForm();
form.Name = "form1";
page.Controls.Add(form1);

MyControl mc = new MyControl();
form.Controls.Add(mc);

StringBuilder sb = new StringBuilder();
StringWriter sw = new StringWriter(sb);
HtmlTextWriter writer = new HtmlTextWriter(sw);

page.RenderControl(writer);

return sb.ToString();

Unfortunately, this gives you more markup than you actually need (since it includes the dummy form). And the ScriptManager will still fail for some arcane reason I haven't puzzled out yet. Honestly, it's a whole lot of trouble and not worth doing; the whole point of generating controls in the code-behind is so that you don't have to fiddle around with the markup, after all.

Solution 5 - C#

override the RenderControl method

protected override void Render(HtmlTextWriter output)
{       
   output.Write("<br>Message from Control : " + Message);       
   output.Write("Showing Custom controls created in reverse" +
                                                    "order");         
   // Render Controls.
   RenderChildren(output);
}

This will give you access to the writer which the HTML will be written to.

You may also want to look into the adaptive control architecture of asp.net adaptive control architecture of asp.net where you can 'shape' the default html output from controls.

Solution 6 - C#

Call its .RenderControl() method.

Solution 7 - C#

You could utilize the HttpServerUtility.Execute Method, available through HttpContext.Current.Server.Execute:

var page = new Page();
var myControl = (MyControl)page.LoadControl("mycontrol.ascx");
myControl.SetSomeProperty = true;
page.Controls.Add(myControl);
var sw = new StringWriter();
HttpContext.Current.Server.Execute(page, sw, preserveForm: false);

The benefit would be that you also trigger the Page_Load event of your user control.

MSDN documentation can be found here: https://msdn.microsoft.com/en-us/library/fb04e8f7(v=vs.110).aspx.

Attributions

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionJon SmockView Question on Stackoverflow
Solution 1 - C#azamsharpView Answer on Stackoverflow
Solution 2 - C#Ben AstonView Answer on Stackoverflow
Solution 3 - C#theJermView Answer on Stackoverflow
Solution 4 - C#ReikimView Answer on Stackoverflow
Solution 5 - C#XianView Answer on Stackoverflow
Solution 6 - C#Joel CoehoornView Answer on Stackoverflow
Solution 7 - C#Carl in 't VeldView Answer on Stackoverflow