1

我的控制器中有两个名称不同的操作,但是当我尝试将数据发送到第二个时,我收到一条错误消息:

当前对控制器类型“HouseholdController”的操作“Index”请求在以下操作方法之间不明确: System.Web.Mvc.ActionResult Find(Int32) on type WhatWorks.Controllers.HouseholdController System.Web.Mvc.ActionResult Index(Int32 ) 在 WhatWorks.Controllers.HouseholdController 类型上

我发现的所有其他问题都与具有相同名称的操作有关。尽管可能有更好的方法来完成我正在尝试的事情,但我无法弄清楚我哪里出错了......

控制器代码

public ActionResult Index(int page = 1)
{
    int pagesize = 10;
    var model = GetDisplay().OrderBy(i => i.familyId);
    return View(model.ToPagedList(page, pagesize));
}

//
// GET: /HouseholdSearch/

public ActionResult Search()
{
    return PartialView("Find");
}

[HttpParamAction]
public ActionResult Find(int Id)
{
    var model = GetDisplay().TakeWhile(m => m.familyId == Id);
    return View("Index", model);
}

局部视图“查找”

@using BootstrapSupport
@model WhatWorks.ViewModels.HouseholdListViewModel

@using (Html.BeginForm()) {
    @Html.ValidationSummary(true)
    <fieldset class="form-horizontal">
        @Html.LabelFor(model => model.familyId, new { @class = "control-label" })
    <div class="controls">
        @Html.TextBoxFor(model => model.familyId, new { @class = "input-mini" })
        @Html.ValidationMessageFor(model => model.familyId, null, new { @class = "help-inline" })
    </div>
    <div>
        <button type="submit" name="Find" class="btn ">Search</button>
    </div>
    </fieldset>
}

HTTPParamAction 代码

using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
using System.Web.Mvc;
using System.Reflection;

public class HttpParamActionAttribute : ActionNameSelectorAttribute
{
    public override bool IsValidName(ControllerContext controllerContext, string actionName, MethodInfo methodInfo)
    {
        if (actionName.Equals(methodInfo.Name, StringComparison.InvariantCultureIgnoreCase))
            return true;

        var request = controllerContext.RequestContext.HttpContext.Request;
        return request[methodInfo.Name] != null;
    }
}
4

2 回答 2

2

尝试将[AcceptVerbs(HttpVerbs.Post)]or[HttpPost]属性添加到您的 find 方法:

[HttpParamAction]
[HttpPost]
public ActionResult Find(int Id)
{
    var model = GetDisplay().TakeWhile(m => m.familyId == Id);
    return View("Index", model);
}
于 2013-10-02T17:38:02.757 回答
1

我已经修改了 BeginForm 以明确引用 Find 操作,这在没有任何额外属性的情况下工作。

我想我只是期待一个更复杂的解决方案!

@using (Html.BeginForm("Find", "Household"))
于 2013-10-02T19:21:08.517 回答