How to use pagedlistpager when passing a viewmodel to my ... | 首都客運時刻表查詢網
![How to use pagedlistpager when passing a viewmodel to my ...](https://i.imgur.com/2oxSoeJ.png)
Itdoesnotmakesensetodoso.YourmethodshouldbeaGET,notaPOSTandyoucantpassacomplexobjectcontainingacollectionpropertytoaGETmethod( ...
![How to use pagedlistpager when passing a viewmodel to my ...](https://i.imgur.com/2oxSoeJ.png)
Ok you can pass a viewmodel back to an action but as previously state it needs to be a GET method so mark your action with the HttpGet attribute. The MVC framework will translate/bind the query string to your view model.
Your controller should look something like this:
[HttpGet] public ActionResult Search(ViewModel viewModel) { \ do something useful here and return ActionResult ..... }You will need to add a page property to your ViewModel and a method allows your .cshtml code to set the page number and returns the viewmodel as an object. MVC will translate the object into a query string for the action link.
Public class ViewModel { // other properties and stuff public int? Page { get; set; } public object ToPagedListParameters(int page) { this.Page = page; return this; } }Then all that is need is a small adjustment to your .cshtml file
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I don't think client-side filtering is an option, due to the fact I'll have a lot of records. @Html.PagedListPager( (IPagedList)@Model.SearchResults, page => ... Read More
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It's going back to your GET action because the Url.Action renders as an <a> tag. An <a> link ALWAYS performs a GET . Read More
How to use pagedlistpager when passing a viewmodel to my ... | 首都客運時刻表查詢網
It does not make sense to do so. Your method should be a GET, not a POST and you cant pass a complex object containing a collection property to a GET method ( ... Read More
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我目前正在使用 Troy Goode's Html.PagedListPager ,但它旨在使用 Url.Action() 发送URL 中的参数。这是一个例子。我不认为客户端过滤是一种选择,因为我会有很多记录 ... Read More
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