How To Generate Thumnail Of Websites Home Page In C#
Feb 12, 2011how could i dynamically generate few thumbnail of my site's home page and shows them in another page.
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Feb 12, 2011how could i dynamically generate few thumbnail of my site's home page and shows them in another page.
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Aug 18, 2010I've got webform routing setup on my asp.net webforms 3.5sp1 project. I would like to have the files for the site in a directory called content including the home page as I would like to run multiple sites using the same system. In MVC there is a blank default page and the home page is in a folder called home. I can't seem to replicate this behaviour using web form routing but would like to. The blank page is always hit first. the route handler is hit second - it recognises that the request is for the home page and sets up the routing page but is not used. the route handler code is simple:
public string VirtualPath { get; private set; }
public IHttpHandler GetHttpHandler(RequestContext
requestContext)
{
string file = requestContext.RouteData.GetRequiredString("File");
string id = requestContext.RouteData.GetRequiredString("Id");
string queryString = "?menuid=" + id;
VirtualPath = "~/" + file;
HttpContext.Current.RewritePath(
string.Concat(
VirtualPath,
queryString));
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(VirtualPath, typeof(Page)) as IHttpHandler;
return page;
}
Is there anyway I can do this?
Update
Here is my global.asax route code:
public static void RegisterRoutes(RouteCollection routes)
{
Domain.RepositoryFactory repo = new RepositoryFactory();
foreach (var x in repo.MenuRepository.GetAllEnabledGetMenus())
{
if (string.IsNullOrEmpty(x.Url))
{
//add default
System.Web.Routing.RouteTable.Routes.Add(
new Route("Default.aspx",
new RouteValueDictionary(new { File = x.FileName,
Id = x.Id.ToString() }),
new CoreRouteHandler()));
}
else
{
string url = x.Url;
if(x.Url.StartsWith("/"))
{
url = url.Remove(0, 1);
}
System.Web.Routing.RouteTable.Routes.Add(
new System.Web.Routing.Route(url,
new RouteValueDictionary(new {File = x.FileName,
Id = x.Id.ToString()}),
new CoreRouteHandler()));
}
}
}