Question

我是MVC的新手，当我尝试通过URL传递多个参数时，我正在努力解决路由问题。

来自包含网址的网页：/PersonCAFDetail/Index/3?memberid=4

...我正在尝试将Html.ActionLink设置为指向Create操作，使得id = 3且memberid = 4.

阅读了许多类似的帖子后，似乎以下内容应该有效：

@Html.ActionLink("Create New", "Create", null, new { memberid = "memberid" })

但是，这会导致创建URL如下：

<a href="/PersonCAFDetail/Create/3" memberid="memberid">Create New</a>

我的路线设置为：

    routes.MapRoute(
        name: "PersonCAFDetail",
        url: "PersonCAFDetail/Create/{id}/{memberid}",
        defaults: new { controller = "PersonCAFDetail", action = "Create", id = "@\d+", memberid = @"\d+" }                        
        );

控制器接受以下两个参数：

 public ActionResult Create(int id, int memberid)
        {
            int cafID = id;
            int personID = memberid;
            ViewBag.detailTypeID = new SelectList(db.tCAFDetailTypes, "detailTypeID", "detailType");
            ViewBag.cafID = new SelectList(db.tFamilyCAFs, "cafID", "issues");
            ViewBag.personID = new SelectList(db.tPersons, "personID", "forename");
            return View();
        }

任何帮助表示感谢。

-------编辑模型----------

namespace WhatWorks.Models
{
    using System;
    using System.Collections.Generic;
    using System.Linq;
    using System.Web;
    using System.Web.Mvc;
    using System.ComponentModel.DataAnnotations;

    public partial class tPersonCAFDetail
    {
        [Key, HiddenInput(DisplayValue=false)]
        public int cafID { get; set; }

        [Key, HiddenInput(DisplayValue = false)]
        public int personID { get; set; }

        [Key, HiddenInput(DisplayValue = false)]
        public int detailTypeID { get; set; }

        [Required, DataType(DataType.MultilineText)]
        public string notes { get; set; }


        public string FullName
        {
            get
            {
                return tPerson.forename + " " + tPerson.surname;
            }
        }

        public virtual tCAFDetailType tCAFDetailType { get; set; }
        public virtual tFamilyCAF tFamilyCAF { get; set; }
        public virtual tPerson tPerson { get; set; }
    }
}

Answer 1

最后，您需要将两个参数传递给视图：

指数行动：

public ActionResult Index(int id, int memberid)
{
    ...
    ViewBag.cafID = id;
    ViewBag.personID = memberid;
    return View();
}

Index.cshtml

@Html.ActionLink("Create New", "Create", "PersonCAFDetail", new { id=ViewBag.cafID , memberid =ViewBag.personID}, null)

并检查您的路线语法... id = @“\ d +”

 routes.MapRoute(
    name: "PersonCAFDetail",
    url: "PersonCAFDetail/Create/{id}/{memberid}",
    defaults: new { controller = "PersonCAFDetail", action = "Create", id = @"\d+", memberid = @"\d+" }                        
    );

Answer 2

Html.ActionLink(string, string, object, object)

..是你正在使用的。这些参数如下：

Html.ActionLink(<link text>, <action name>, <route values>, <html attributes>

您将数据放入属性参数中，这自然会使它们成为您链接的属性（而不是路由值）。

用法示例：

@Html.ActionLink("Create new", "Create", new { id = Model.cafID, memberid = Model.personID }, null);

Answer 3

因为您的Url.Action不起作用是＆amp; url中的char已编码，因此您必须使用

@Html.Raw(Html.ActionLink("Create New", "Create", "PersonCAFDetail", new { id=ViewBag.cafID , memberid =ViewBag.personID}, null))

现在，它正在运作;）

Answer 4

您应该创建一个具有重载方法的ActionLink

MvcHtmlString HtmlHelper.ActionLink(
    string linkText, 
    string actionName, 
    string controllerName, 
    object routeValues, 
    object htmlAttributes
)

您的情况：

@Html.ActionLink("Create New", "Create", PersonCAFDetail, new { id = "id", memberid = "memberid" })

将多个参数从url传递到html.actionlink

4 个答案: