C# 如何在 ASP.NET MVC 中进行分页？

How do I do pagination in ASP.NET MVC?

提问by Spoike

What is the most preferred and easiest way to do pagination in ASP.NET MVC? I.e. what is the easiest way to break up a list into several browsable pages.

在 ASP.NET MVC 中进行分页的最首选和最简单的方法是什么？即什么是将列表分解为几个可浏览页面的最简单方法。

As an example lets say I get a list of elements from a database/gateway/repository like this:

例如，假设我从数据库/网关/存储库中获取元素列表，如下所示：

public ActionResult ListMyItems()
{
    List<Item> list = ItemDB.GetListOfItems();
    ViewData["ItemList"] = list;

    return View();
}

For simplicity's sake I'd like to specify just a page number for my action as parameter. Like this:

为简单起见，我只想为我的操作指定一个页码作为参数。像这样：

public ActionResult ListMyItems(int page)
{
   //...
}

采纳答案by Marc Gravell

Well, what is the data source? Your action could take a few defaulted arguments, i.e.

那么，数据源是什么？您的操作可能需要一些默认参数，即

ActionResult Search(string query, int startIndex, int pageSize) {...}

defaulted in the routes setup so that startIndex is 0 and pageSize is (say) 20:

在路由设置中默认设置为 startIndex 为 0，pageSize 为（例如）20：

        routes.MapRoute("Search", "Search/{query}/{startIndex}",
                        new
                        {
                            controller = "Home", action = "Search",
                            startIndex = 0, pageSize = 20
                        });

To split the feed, you can use LINQ quite easily:

要拆分提要，您可以很容易地使用 LINQ：

var page = source.Skip(startIndex).Take(pageSize);

(or do a multiplication if you use "pageNumber" rather than "startIndex")

（或者，如果您使用“pageNumber”而不是“startIndex”，则进行乘法运算）

With LINQ-toSQL, EF, etc - this should "compose" down to the database, too.

使用 LINQ-toSQL、EF 等 - 这也应该“组合”到数据库中。

You should then be able to use action-links to the next page (etc):

然后，您应该能够使用操作链接到下一页（等）：

<%=Html.ActionLink("next page", "Search", new {
                query, startIndex = startIndex + pageSize, pageSize }) %>

回答by Oliver

I had the same problem and found a very elegant solution for a Pager Class from

我遇到了同样的问题，并为 Pager Class 找到了一个非常优雅的解决方案

http://blogs.taiga.nl/martijn/2008/08/27/paging-with-aspnet-mvc/

http://blogs.taiga.nl/martijn/2008/08/27/paging-with-aspnet-mvc/

In your controller the call looks like:

在您的控制器中，调用如下所示：

return View(partnerList.ToPagedList(currentPageIndex, pageSize));

and in your view:

在您看来：

<div class="pager">
    Seite: <%= Html.Pager(ViewData.Model.PageSize, 
                          ViewData.Model.PageNumber,
                          ViewData.Model.TotalItemCount)%>
</div>

回答by dav_i

I wanted to cover a simple way of doing this with the front end too:

我也想介绍一种在前端执行此操作的简单方法：

Controller:

控制器：

public ActionResult Index(int page = 0)
{
    const int PageSize = 3; // you can always do something more elegant to set this

    var count = this.dataSource.Count();

    var data = this.dataSource.Skip(page * PageSize).Take(PageSize).ToList();

    this.ViewBag.MaxPage = (count / PageSize) - (count % PageSize == 0 ? 1 : 0);

    this.ViewBag.Page = page;

    return this.View(data);
}

View:

看法：

@* rest of file with view *@

@if (ViewBag.Page > 0)
{
    <a href="@Url.Action("Index", new { page = ViewBag.Page - 1 })" 
       class="btn btn-default">
        &laquo; Prev
    </a>
}
@if (ViewBag.Page < ViewBag.MaxPage)
{
    <a href="@Url.Action("Index", new { page = ViewBag.Page + 1 })" 
       class="btn btn-default">
        Next &raquo;
    </a>
}

回答by DotNetLover

Controller

控制器

 [HttpGet]
    public async Task<ActionResult> Index(int page =1)
    {
        if (page < 0 || page ==0 )
        {
            page = 1;
        }
        int pageSize = 5;
        int totalPage = 0;
        int totalRecord = 0;
        BusinessLayer bll = new BusinessLayer();
        MatchModel matchmodel = new MatchModel();
        matchmodel.GetMatchList = bll.GetMatchCore(page, pageSize, out totalRecord, out totalPage);
        ViewBag.dbCount = totalPage;
        return View(matchmodel);
    }

BusinessLogic

商业逻辑

  public List<Match> GetMatchCore(int page, int pageSize, out int totalRecord, out int totalPage)
    {
        SignalRDataContext db = new SignalRDataContext();
        var query = new List<Match>();
        totalRecord = db.Matches.Count();
        totalPage = (totalRecord / pageSize) + ((totalRecord % pageSize) > 0 ? 1 : 0);
        query = db.Matches.OrderBy(a => a.QuestionID).Skip(((page - 1) * pageSize)).Take(pageSize).ToList();
        return query;
    }

View for displaying total page count

用于显示总页数的视图

 if (ViewBag.dbCount != null)
    {
        for (int i = 1; i <= ViewBag.dbCount; i++)
        {
            <ul class="pagination">
                <li>@Html.ActionLink(@i.ToString(), "Index", "Grid", new { page = @i },null)</li> 
            </ul>
        }
    }

回答by LENG UNG

I think the easiest way to create pagination in ASP.NET MVC application is using PagedList library.

我认为在 ASP.NET MVC 应用程序中创建分页的最简单方法是使用 PagedList 库。

There is a complete example in following github repository. Hope it would help.

以下github存储库中有一个完整的示例。希望它会有所帮助。

public class ProductController : Controller
{
    public object Index(int? page)
    {
        var list = ItemDB.GetListOfItems();

        var pageNumber = page ?? 1; 
        var onePageOfItem = list.ToPagedList(pageNumber, 25); // will only contain 25 items max because of the pageSize

        ViewBag.onePageOfItem = onePageOfProducts;
        return View();
    }
}

Demo Link: http://ajaxpagination.azurewebsites.net/

演示链接：http: //ajaxpagination.azurewebsites.net/

Source Code: https://github.com/ungleng/SimpleAjaxPagedListAndSearchMVC5

源代码：https: //github.com/ungleng/SimpleAjaxPagedListAndSearchMVC5

回答by Kishu

Entity

实体

public class PageEntity
{
    public int Page { get; set; }
    public string Class { get; set; }
}

public class Pagination
{
    public List<PageEntity> Pages { get; set; }
    public int Next { get; set; }
    public int Previous { get; set; }
    public string NextClass { get; set; }
    public string PreviousClass { get; set; }
    public bool Display { get; set; }
    public string Query { get; set; }
}

HTML

HTML

<nav>
    <div class="navigation" style="text-align: center">
        <ul class="pagination">
            <li class="page-item @Model.NextClass"><a class="page-link" href="?page=@(@[email protected])">&laquo;</a></li>
            @foreach (var item in @Model.Pages)
            {
                <li class="page-item @item.Class"><a class="page-link" href="?page=@([email protected])">@item.Page</a></li>
            }
            <li class="page-item @Model.NextClass"><a class="page-link" href="?page=@(@[email protected])">&raquo;</a></li>
        </ul>
    </div>
 </nav>

Paging Logic

分页逻辑

public Pagination GetCategoryPaging(int currentPage, int recordCount, string query)
{
    string pageClass = string.Empty; int pageSize = 10, innerCount = 5;

    Pagination pagination = new Pagination();
    pagination.Pages = new List<PageEntity>();
    pagination.Next = currentPage + 1;
    pagination.Previous = ((currentPage - 1) > 0) ? (currentPage - 1) : 1;
    pagination.Query = query;

    int totalPages = ((int)recordCount % pageSize) == 0 ? (int)recordCount / pageSize : (int)recordCount / pageSize + 1;

    int loopStart = 1, loopCount = 1;

    if ((currentPage - 2) > 0)
    {
        loopStart = (currentPage - 2);
    }

    for (int i = loopStart; i <= totalPages; i++)
    {
        pagination.Pages.Add(new PageEntity { Page = i, Class = string.Empty });

        if (loopCount == innerCount)
        { break; }

        loopCount++;
    }

    if (totalPages <= innerCount)
    {
        pagination.PreviousClass = "disabled";
    }

    foreach (var item in pagination.Pages.Where(x => x.Page == currentPage))
    {
        item.Class = "active";
    }

    if (pagination.Pages.Count() <= 1)
    {
        pagination.Display = false;
    }

    return pagination;
}

Using Controller

使用控制器

public ActionResult GetPages()
{
    int currentPage = 1; string search = string.Empty;
    if (!string.IsNullOrEmpty(Request.QueryString["page"]))
    {
        currentPage = Convert.ToInt32(Request.QueryString["page"]);
    }

    if (!string.IsNullOrEmpty(Request.QueryString["q"]))
    {
        search = "&q=" + Request.QueryString["q"];
    }
    /* to be Fetched from database using count */
    int recordCount = 100;

    Place place = new Place();
    Pagination pagination = place.GetCategoryPaging(currentPage, recordCount, search);

    return PartialView("Controls/_Pagination", pagination);
}

回答by Prashant vishwakarma

public ActionResult Paging(int? pageno,bool? fwd,bool? bwd)        
{
    if(pageno!=null)
     {
       Session["currentpage"] = pageno;
     }

    using (HatronEntities DB = new HatronEntities())
    {
        if(fwd!=null && (bool)fwd)
        {
            pageno = Convert.ToInt32(Session["currentpage"]) + 1;
            Session["currentpage"] = pageno;
        }
        if (bwd != null && (bool)bwd)
        {
            pageno = Convert.ToInt32(Session["currentpage"]) - 1;
            Session["currentpage"] = pageno;
        }
        if (pageno==null)
        {
            pageno = 1;
        }
        if(pageno<0)
        {
            pageno = 1;
        }
        int total = DB.EmployeePromotion(0, 0, 0).Count();
        int  totalPage = (int)Math.Ceiling((double)total / 20);
        ViewBag.pages = totalPage;
        if (pageno > totalPage)
        {
            pageno = totalPage;
        }
        return View (DB.EmployeePromotion(0,0,0).Skip(GetSkip((int)pageno,20)).Take(20).ToList());     
    }
}

private static int GetSkip(int pageIndex, int take)
{
    return (pageIndex - 1) * take;
}

@model IEnumerable<EmployeePromotion_Result>
@{
  Layout = null;
}

 <!DOCTYPE html>

 <html>
 <head>
    <meta name="viewport" content="width=device-width" />
    <title>Paging</title>
  </head>
  <body>
 <div> 
    <table border="1">
        @foreach (var itm in Model)
        {
 <tr>
   <td>@itm.District</td>
   <td>@itm.employee</td>
   <td>@itm.PromotionTo</td>
 </tr>
        }
    </table>
    <a href="@Url.Action("Paging", "Home",new { pageno=1 })">First  page</a> 
    <a href="@Url.Action("Paging", "Home", new { bwd =true })"><<</a> 
    @for(int itmp =1; itmp< Convert.ToInt32(ViewBag.pages)+1;itmp++)
   {
       <a href="@Url.Action("Paging", "Home",new { pageno=itmp   })">@itmp.ToString()</a>
   }
    <a href="@Url.Action("Paging", "Home", new { fwd = true })">>></a> 
    <a href="@Url.Action("Paging", "Home", new { pageno =                                                                               Convert.ToInt32(ViewBag.pages) })">Last page</a> 
</div>
   </body>
  </html>

回答by Aja Enyi

Here's a linkthat helped me with this.

这是一个帮助我解决这个问题的链接。

It uses PagedList.MVC NuGet package. I'll try to summarize the steps

它使用 PagedList.MVC NuGet 包。我会尝试总结步骤

1. Install the PagedList.MVC NuGet package

2. Build project

3. Addusing PagedList;to the controller

4. Modify your action to set pagepublic ActionResult ListMyItems(int? page) { List list = ItemDB.GetListOfItems(); int pageSize = 3; int pageNumber = (page ?? 1); return View(list.ToPagedList(pageNumber, pageSize)); }

5. Add paging links to the bottom of your view@*Your existing view*@ Page @(Model.PageCount < Model.PageNumber ? 0 : Model.PageNumber) of @Model.PageCount @Html.PagedListPager(Model, page => Url.Action("Index", new { page, sortOrder = ViewBag.CurrentSort, currentFilter = ViewBag.CurrentFilter }))

1. 安装 PagedList.MVC NuGet 包

2. 构建项目

3. 添加using PagedList;到控制器

4. 修改您的操作以设置页面public ActionResult ListMyItems(int? page) { List list = ItemDB.GetListOfItems(); int pageSize = 3; int pageNumber = (page ?? 1); return View(list.ToPagedList(pageNumber, pageSize)); }

5. 在视图底部添加分页链接@*Your existing view*@ Page @(Model.PageCount < Model.PageNumber ? 0 : Model.PageNumber) of @Model.PageCount @Html.PagedListPager(Model, page => Url.Action("Index", new { page, sortOrder = ViewBag.CurrentSort, currentFilter = ViewBag.CurrentFilter }))