Question

嗨，当使用PagedList与ViewModel（与域模型完美配合）时，你真的需要你帮助解决问题。

我是MVC初学者，这是我到目前为止所尝试过的。

/ **********我有一个型号*********** /

public class Property
{        
    public int ID { get; set; }
    public int UserID { get; set; }
    public DateTime Created { get; set; }
    public DateTime Edited { get; set; }
}

从中创建ViewModel（暂时只使用一个Domain模型从其他Domain模型添加更多属性

public class SearchResultsVM
{

    public PagedList.IPagedList<Property> BasicDetails { get; set; }

}

控制器操作方法：

public ActionResult SearchResults(string sortOrder, string SearchKeyword, string currentFilter, int? page)
{
    ViewBag.CurrentSort = sortOrder;
    ViewBag.TitleSortParm = String.IsNullOrEmpty(sortOrder) ? "title_desc" : "";
    ViewBag.DateSortParm = sortOrder == "Date" ? "date_desc" : "Date";
    if (SearchKeyword != null){page = 1;}
    else{SearchKeyword = currentFilter;}
    ViewBag.CurrentFilter = SearchKeyword;            
    var sr = from s in db.Property
                          select s;          
    if (!String.IsNullOrEmpty(SearchKeyword))
    {sr = db.Property.Where(s =>   s.PropertyTitle.ToUpper().Contains(SearchKeyword.ToUpper()));}
    switch (sortOrder)
    {
        case "title_desc":
            sr = sr.OrderByDescending(s => s.PropertyTitle);
            break;
        case "Date":
            sr = sr.OrderBy(s => s.Created);
            break;
        case "date_desc":
            sr = sr.OrderByDescending(s => s.Created);
            break;
        default:
            sr = sr.OrderBy(s => s.PropertyTitle);
            break;
    }
    SearchResultsVM srVM = new SearchResultsVM();
    int pageSize = 10;
    int pageNumber = (page ?? 1);
    srVM.BasicDetails = sr.ToList().ToPagedList(pageNumber,pageSize);
    return View(srVM);
}

我的观点

在View中访问ViewModel时我没有获得Property Domain Model的任何属性

@model PagedList.IPagedList<GH_Final.ViewModel.SearchResultsVM>
@using PagedList;
@using PagedList.Mvc;

@using (Html.BeginForm("SearchResults", "Home", FormMethod.Get))
{
<p>
    Find by name: @Html.TextBox("SearchKeyword", ViewBag.CurrentFilter as string)
    <input type="submit" value="Search" />
</p>
}


@foreach (var item in Model)
{
    <tr>
        <td>
            @Html.DisplayFor(modelItem => item.created)
        </td>
        <td>
            @Html.DisplayFor(modelItem => item.BasicDetails.Created)
        </td>
        <td>
            @Html.DisplayFor(modelItem => item.BasicDetails.PropertyDesc)
        </td>

Answer 1

而不是：

@model PagedList.IPagedList<GH_Final.ViewModel.SearchResultsVM>

这样做：

@model GH_Final.ViewModel.SearchResultsVM

而不是：

@foreach (var item in Model)
{
    <tr>
        <td>
            @Html.DisplayFor(modelItem => item.created)
        </td>
        <td>
            @Html.DisplayFor(modelItem => item.BasicDetails.Created)
        </td>
        <td>
            @Html.DisplayFor(modelItem => item.BasicDetails.PropertyDesc)
        </td>

应该是：

@foreach (var item in Model.BasicDetails)
{
    <tr>
        <td>
            @Html.DisplayFor(modelItem => item.Created)
        </td>
        <td>
            @Html.DisplayFor(modelItem => item.Edited)
        </td>
        <td>
            @Html.DisplayFor(modelItem => item.PropertyDesc)
        </td>
    <tr>
}

Answer 2

您还应将表达式设为：

@Html.DisplayFor(modelItem => item.First().Created)
@Html.DisplayFor(modelItem => item.First().Edited)
@Html.DisplayFor(modelItem => item.First().PropertyDesc)

将PagedList与ViewModel一起使用

我的观点

2 个答案: