asp.net mvc从GET form post出站路由

我有一条路线如下所示：

new { controller = "ControllerName", action = "PhotoGallery", slug = "photo-gallery", filtertype = UrlParameter.Optional, filtervalue = UrlParameter.Optional, sku = UrlParameter.Optional}

  <%using(Html.BeginForm("PhotoGallery", "ControllerName", FormMethod.Get)) {%>
  <%:Html.Hidden("filtertype", "1")%>
  <%:Html.DropDownList("filtervalue", ViewData["Designers"] as SelectList, "Photos by designer", new { onchange = "selectJump(this)" })%>
  <%}%>

www.site.com/photo-gallery/1/selectedvalue

我有一张这样的表格：

new { controller = "ControllerName", action = "PhotoGallery", slug = "photo-gallery", filtertype = UrlParameter.Optional, filtervalue = UrlParameter.Optional, sku = UrlParameter.Optional}

  <%using(Html.BeginForm("PhotoGallery", "ControllerName", FormMethod.Get)) {%>
  <%:Html.Hidden("filtertype", "1")%>
  <%:Html.DropDownList("filtervalue", ViewData["Designers"] as SelectList, "Photos by designer", new { onchange = "selectJump(this)" })%>
  <%}%>

www.site.com/photo-gallery/1/selectedvalue

而不是像“

---

谢谢！

如果您只有一些参数，Mvc将创建一个查询字符串类型的Url，除非它找到一个完全匹配的Url

建议您将需要以下内容：

routes.Add("testRoute", 
new Route("/{action}/{controller}/{slug}/{filtertype}/{filtervalue}",
new { controller = "ControllerName", action = "PhotoGallery", slug = "photo-gallery", filtertype = UrlParameter.Optional, filtervalue = UrlParameter.Optional}
);

---

确保这是在您现有路线之前进行的

谢谢您的回复！我最终使用了PRG模式以获得所需的效果。我的路线太复杂了，不允许有这样的一般规则。