6

@Html.Pager来自MvcPaging 2.0的Html 助手。有.Options(o => o.RouteValues(object RouteValues))它可以将Model返回给Controller，但是MvcPaging需要在他所在的View中填充这个helper IPagedList<model>。这是生成表和分页的Model。实现 mvcpaging 2.0 的最佳方法是什么？使用 SearchModel 进行搜索并使用模型显示结果？

例子：

楷模：

public class SearchModel
{
    public string FirstName { get; set; }
    public string LastName { get; set; }
}

public class Person
{
    [Key]
    public int Id { get; set; }
    public string FirstName { get; set; }
    public string LastName { get; set; }
    public DateTime Dob { get; set; }
    public string City { get; set; }
}

查看： Index.cshtml

@using (Ajax.BeginForm("Search", "SearchPerson", new AjaxOptions
{
    HttpMethod = "GET",
    InsertionMode = InsertionMode.Replace,
    UpdateTargetId = "main_search_result_table_id"
}))
{    
    @Html.TextBoxFor(m => m.FirstName)
    @Html.TextBoxFor(m => m.LastName)
    <input type="submit" value="Search"/>
}
 <div id="main_search_result_table_id">
      @{Html.RenderPartial("_InitPartialEmpty");}
 </div>

_ResultPartial.cshtml

@using MvcPaging
@model IPagedList<Models.Person>

<table>
@foreach (var p in Model)
{
<tr>
    <td>@p.FirstName</td>
    <td>@p.LastName</td>
    <td>@p.Dob</td>
    <td>@p.City</td>
</tr>
}
<table>

 @Html.Pager(Model.PageSize, Model.PageNumber,
                 Model.TotalItemCount,  new AjaxOptions 
 { 
     UpdateTargetId = "main_search_result_table_id"
 }).Options(o => o.RouteValues(Model)) //==> IPagedList<Models.Person>

控制器

public ActionResult SearchPerson(int? page,SearchModel person)
{
    List<Person> result= adapter.GetPersons(person);

    int currentPageIndex = page.HasValue ? page.Value - 1 : 0;

    return PartialView("_ResultPartial", 
                result.ToPagedList(currentPageIndex, 10, result.Count()));
}

问题是如何实现 MvcPaging2.0 使用模型进行搜索？或者是否有另一种方式，更好的方式来进行复杂的搜索而​​不使用模型来传输数据查询？有什么想法吗？

我正在使用MvcPaging 2.0。,文档

编辑：*

感谢达林的回答，但我设法把它拉成这样：

*_ResultPartial.cshtml*

@Html.Pager(Model.PageSize, Model.PageNumber,
             Model.TotalItemCount,  new AjaxOptions 
{ 
 UpdateTargetId = "main_search_result_table_id"
 }).Options(o => o.Action("AjaxPaging"))

控制器

public ActionResult SearchPerson(int? page,SearchModel person)
{
    IQueryable<Person> query= adapter.GetPersons(person);

    Session["SearchQuery"] = query;

    int currentPageIndex = page.HasValue ? page.Value - 1 : 0;

    List<Person> persons = query.ToList();

    return PartialView("_ResultPartial", 
                persons.ToPagedList(currentPageIndex, 10, persons.Count()));
}


public ActionResult AjaxPaging(int? page)
{
    IQueryable<Person> query = Session["SearchQuery"] as IQueryable<Person>;

    int currentPageIndex = page.HasValue ? page.Value - 1 : 0;

    List<Person> persons = query.ToList();

    return PartialView("_ResultPartial", 
                persons.ToPagedList(currentPageIndex, 10, persons.Count()));
}
4

1 回答 1

5

您可以编写一个自定义扩展方法，该方法将收集所有查询字符串参数并将它们添加到页面链接中，以及所有 currentPage、pageNumber、totalItemCount 等......

public static class PagerOptionsBuilderExtensions
{
    public static PagerOptionsBuilder AddFromQueryString(
        this PagerOptionsBuilder builder, 
        HttpRequestBase request
    )
    {
        foreach (string item in request.QueryString)
        {
            builder.AddRouteValue(item, request.QueryString[item]);
        }
        return builder;
    }
}

接着：

.Options(o => o.RouteValues(Model).AddFromQueryString(Request))
于 2012-08-20T09:54:04.387 回答