ASP.NET MVC pagedlist loose checkboxlist filter











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I'm working on an Asp MVC application that contains a grid of data that I can filter with a checkboxlist.I'm using PagedList to display the data. My filter works well on the first page, but if I click on the 2nd page, the filter is cancelled. I don't have this problem with filter using dropdownlist(it's not in the following example).



My view looks like this:



<div class="wrapper">
<nav id="sidebar">
@using (Html.BeginForm("Index", "Missions", FormMethod.Get))
{
@Html.EditorFor(x => x.Decision)
}
</nav>
<div id="content" class="container">
<table class="table table-bordered">
<tr>
<th class="col-md-2">
Decision
</th>

</tr>
@foreach (var item in Model.OnePageOfMissions)
{
<tr>
<td class="col-md-1">
@Html.DisplayFor(modelItem => item.decision)
</td>
</tr>
}
@Html.PagedListPager((IPagedList)Model.OnePageOfMissions, page => Url.Action("Index", new { page, Decision = Model.Decision}))
</table>
</div>




I created a template to display checkboxlist like this:



<div class="form-check row">
@Html.HiddenFor(x => x.ID)
@Html.CheckBoxFor(x => x.IsChecked, htmlAttributes: new { onchange = "form.submit();", @class = "form-check-input col-md-2" })
@Html.LabelFor(x => x.Display, Model.Display, htmlAttributes: new { @class = "form-check-label col-md-8" })




My controller looks like this:



 public ActionResult Index(int? page, List<CheckBoxListItem> Decision)
{
IndexViewModel model = new IndexViewModel();
//Display Missions
model.missionsList = db.missions_supportmission.ToList();
//Retrieve parameters
model.Decision = Decision;
//PagedList
var pageNumber = page ?? 1; // if no page was specified in the querystring, default to the first page (1)
var onePageOfMissions = model.missionsList.ToPagedList(pageNumber, 10); // will only contain 10 products max because of the pageSize(equel to 10)
model.OnePageOfMissions = onePageOfMissions;
//Filter
if (model.Decision != null)
{
var selecteddecision = model.Decision.Where(x => x.IsChecked).Select(x => x.ID);
model.OnePageOfMissions = (from m in db.missions_supportmission
join l in db.list_decision
on m.decision equals l.decision_id
where selecteddecision.Contains(l.decision_id)
select m)
.OrderBy(a => a.id)
.ToPagedList(pageNumber, 10);
}
}
}
//Display CheckBox
//Checkboxlist (!important => mettre ce bloc de code après la requête link qui permet de filtrer sur les checkbox)
var allDecisions = db.list_decision.ToList();//returns List<list_decision>
var checkBoxListItems = new List<CheckBoxListItem>(); //nouvelle instance de la classe checkboxlist
model.Decision = checkBoxListItems;
foreach (var decison in allDecisions)
{//On assigne les valeurs "id", "display" et "is checked" à la variable checkboxlistitem
checkBoxListItems.Add(new CheckBoxListItem()
{
ID = decison.decision_id,
Display = decison.name_en,
IsChecked = false //On the add view, no decision are selected by default
});
}

return View(model);
}


I have also a ViewModel:



namespace MissionsDF.Models
{
public class IndexViewModel
{

public IEnumerable<missions_supportmission> missionsList { get; set; }
public List<CheckBoxListItem> Decision { get; set; }
public IndexViewModel()
{
this.Decision = new List<CheckBoxListItem>();
}
public IPagedList<missions_supportmission> OnePageOfMissions { get; set; }

}
}









share|improve this question






















  • You cannot pass a collection of complex objects in a url using new { Decision = Model.Decision} (look at the url it is generating) - you need to pass each property of each item in the collection to make it bind - but that would make no sense because you would probably exceed the query string limit and throw an exception. One option would be to have a (say) string SelectedDecisions property in the view model containing a comma separated string of the selected ID's, and pass that back in the url, and split it in the POST method.
    – Stephen Muecke
    Nov 7 at 20:55










  • I don't see how to do have a property containing a comma separated string of the selected ID's. Can you illustrate your idea?
    – AmélieV
    Nov 8 at 8:59










  • Add that property to your view model (and set its value using if(Decision != null){ var selected = Decision.Where(x => x.IsChecked).Select(x => x.ID); model.SelectedDecisions = String.Join(",", selected); } else { model.SelectedDecisions = selectedDecisions; }, then add a string selectedDecisions parameter to your model and in the view use @Html.PagedListPager((IPagedList)Model.OnePageOfMissions, page => Url.Action("Index", new { page, selectedDecisions = Model.SelectedDecisions}).
    – Stephen Muecke
    Nov 8 at 9:10










  • If you post the form, Decision will have a value, otherwise if you click a page number, then selectedDecisions will have the value. But your new { onchange = "form.submit(); in the CheckBoxFor() is terrible practice (and means the user can only ever click one checkbox) - you should allow the user to make their selections and then click a submit button
    – Stephen Muecke
    Nov 8 at 9:13










  • You will also need to modify the if (model.Decision != null) code so that if it is null (i.e. user selected a page number), then you would use String.Split(...) to set your var selecteddecision variable
    – Stephen Muecke
    Nov 8 at 9:20















up vote
0
down vote

favorite












I'm working on an Asp MVC application that contains a grid of data that I can filter with a checkboxlist.I'm using PagedList to display the data. My filter works well on the first page, but if I click on the 2nd page, the filter is cancelled. I don't have this problem with filter using dropdownlist(it's not in the following example).



My view looks like this:



<div class="wrapper">
<nav id="sidebar">
@using (Html.BeginForm("Index", "Missions", FormMethod.Get))
{
@Html.EditorFor(x => x.Decision)
}
</nav>
<div id="content" class="container">
<table class="table table-bordered">
<tr>
<th class="col-md-2">
Decision
</th>

</tr>
@foreach (var item in Model.OnePageOfMissions)
{
<tr>
<td class="col-md-1">
@Html.DisplayFor(modelItem => item.decision)
</td>
</tr>
}
@Html.PagedListPager((IPagedList)Model.OnePageOfMissions, page => Url.Action("Index", new { page, Decision = Model.Decision}))
</table>
</div>




I created a template to display checkboxlist like this:



<div class="form-check row">
@Html.HiddenFor(x => x.ID)
@Html.CheckBoxFor(x => x.IsChecked, htmlAttributes: new { onchange = "form.submit();", @class = "form-check-input col-md-2" })
@Html.LabelFor(x => x.Display, Model.Display, htmlAttributes: new { @class = "form-check-label col-md-8" })




My controller looks like this:



 public ActionResult Index(int? page, List<CheckBoxListItem> Decision)
{
IndexViewModel model = new IndexViewModel();
//Display Missions
model.missionsList = db.missions_supportmission.ToList();
//Retrieve parameters
model.Decision = Decision;
//PagedList
var pageNumber = page ?? 1; // if no page was specified in the querystring, default to the first page (1)
var onePageOfMissions = model.missionsList.ToPagedList(pageNumber, 10); // will only contain 10 products max because of the pageSize(equel to 10)
model.OnePageOfMissions = onePageOfMissions;
//Filter
if (model.Decision != null)
{
var selecteddecision = model.Decision.Where(x => x.IsChecked).Select(x => x.ID);
model.OnePageOfMissions = (from m in db.missions_supportmission
join l in db.list_decision
on m.decision equals l.decision_id
where selecteddecision.Contains(l.decision_id)
select m)
.OrderBy(a => a.id)
.ToPagedList(pageNumber, 10);
}
}
}
//Display CheckBox
//Checkboxlist (!important => mettre ce bloc de code après la requête link qui permet de filtrer sur les checkbox)
var allDecisions = db.list_decision.ToList();//returns List<list_decision>
var checkBoxListItems = new List<CheckBoxListItem>(); //nouvelle instance de la classe checkboxlist
model.Decision = checkBoxListItems;
foreach (var decison in allDecisions)
{//On assigne les valeurs "id", "display" et "is checked" à la variable checkboxlistitem
checkBoxListItems.Add(new CheckBoxListItem()
{
ID = decison.decision_id,
Display = decison.name_en,
IsChecked = false //On the add view, no decision are selected by default
});
}

return View(model);
}


I have also a ViewModel:



namespace MissionsDF.Models
{
public class IndexViewModel
{

public IEnumerable<missions_supportmission> missionsList { get; set; }
public List<CheckBoxListItem> Decision { get; set; }
public IndexViewModel()
{
this.Decision = new List<CheckBoxListItem>();
}
public IPagedList<missions_supportmission> OnePageOfMissions { get; set; }

}
}









share|improve this question






















  • You cannot pass a collection of complex objects in a url using new { Decision = Model.Decision} (look at the url it is generating) - you need to pass each property of each item in the collection to make it bind - but that would make no sense because you would probably exceed the query string limit and throw an exception. One option would be to have a (say) string SelectedDecisions property in the view model containing a comma separated string of the selected ID's, and pass that back in the url, and split it in the POST method.
    – Stephen Muecke
    Nov 7 at 20:55










  • I don't see how to do have a property containing a comma separated string of the selected ID's. Can you illustrate your idea?
    – AmélieV
    Nov 8 at 8:59










  • Add that property to your view model (and set its value using if(Decision != null){ var selected = Decision.Where(x => x.IsChecked).Select(x => x.ID); model.SelectedDecisions = String.Join(",", selected); } else { model.SelectedDecisions = selectedDecisions; }, then add a string selectedDecisions parameter to your model and in the view use @Html.PagedListPager((IPagedList)Model.OnePageOfMissions, page => Url.Action("Index", new { page, selectedDecisions = Model.SelectedDecisions}).
    – Stephen Muecke
    Nov 8 at 9:10










  • If you post the form, Decision will have a value, otherwise if you click a page number, then selectedDecisions will have the value. But your new { onchange = "form.submit(); in the CheckBoxFor() is terrible practice (and means the user can only ever click one checkbox) - you should allow the user to make their selections and then click a submit button
    – Stephen Muecke
    Nov 8 at 9:13










  • You will also need to modify the if (model.Decision != null) code so that if it is null (i.e. user selected a page number), then you would use String.Split(...) to set your var selecteddecision variable
    – Stephen Muecke
    Nov 8 at 9:20













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I'm working on an Asp MVC application that contains a grid of data that I can filter with a checkboxlist.I'm using PagedList to display the data. My filter works well on the first page, but if I click on the 2nd page, the filter is cancelled. I don't have this problem with filter using dropdownlist(it's not in the following example).



My view looks like this:



<div class="wrapper">
<nav id="sidebar">
@using (Html.BeginForm("Index", "Missions", FormMethod.Get))
{
@Html.EditorFor(x => x.Decision)
}
</nav>
<div id="content" class="container">
<table class="table table-bordered">
<tr>
<th class="col-md-2">
Decision
</th>

</tr>
@foreach (var item in Model.OnePageOfMissions)
{
<tr>
<td class="col-md-1">
@Html.DisplayFor(modelItem => item.decision)
</td>
</tr>
}
@Html.PagedListPager((IPagedList)Model.OnePageOfMissions, page => Url.Action("Index", new { page, Decision = Model.Decision}))
</table>
</div>




I created a template to display checkboxlist like this:



<div class="form-check row">
@Html.HiddenFor(x => x.ID)
@Html.CheckBoxFor(x => x.IsChecked, htmlAttributes: new { onchange = "form.submit();", @class = "form-check-input col-md-2" })
@Html.LabelFor(x => x.Display, Model.Display, htmlAttributes: new { @class = "form-check-label col-md-8" })




My controller looks like this:



 public ActionResult Index(int? page, List<CheckBoxListItem> Decision)
{
IndexViewModel model = new IndexViewModel();
//Display Missions
model.missionsList = db.missions_supportmission.ToList();
//Retrieve parameters
model.Decision = Decision;
//PagedList
var pageNumber = page ?? 1; // if no page was specified in the querystring, default to the first page (1)
var onePageOfMissions = model.missionsList.ToPagedList(pageNumber, 10); // will only contain 10 products max because of the pageSize(equel to 10)
model.OnePageOfMissions = onePageOfMissions;
//Filter
if (model.Decision != null)
{
var selecteddecision = model.Decision.Where(x => x.IsChecked).Select(x => x.ID);
model.OnePageOfMissions = (from m in db.missions_supportmission
join l in db.list_decision
on m.decision equals l.decision_id
where selecteddecision.Contains(l.decision_id)
select m)
.OrderBy(a => a.id)
.ToPagedList(pageNumber, 10);
}
}
}
//Display CheckBox
//Checkboxlist (!important => mettre ce bloc de code après la requête link qui permet de filtrer sur les checkbox)
var allDecisions = db.list_decision.ToList();//returns List<list_decision>
var checkBoxListItems = new List<CheckBoxListItem>(); //nouvelle instance de la classe checkboxlist
model.Decision = checkBoxListItems;
foreach (var decison in allDecisions)
{//On assigne les valeurs "id", "display" et "is checked" à la variable checkboxlistitem
checkBoxListItems.Add(new CheckBoxListItem()
{
ID = decison.decision_id,
Display = decison.name_en,
IsChecked = false //On the add view, no decision are selected by default
});
}

return View(model);
}


I have also a ViewModel:



namespace MissionsDF.Models
{
public class IndexViewModel
{

public IEnumerable<missions_supportmission> missionsList { get; set; }
public List<CheckBoxListItem> Decision { get; set; }
public IndexViewModel()
{
this.Decision = new List<CheckBoxListItem>();
}
public IPagedList<missions_supportmission> OnePageOfMissions { get; set; }

}
}









share|improve this question













I'm working on an Asp MVC application that contains a grid of data that I can filter with a checkboxlist.I'm using PagedList to display the data. My filter works well on the first page, but if I click on the 2nd page, the filter is cancelled. I don't have this problem with filter using dropdownlist(it's not in the following example).



My view looks like this:



<div class="wrapper">
<nav id="sidebar">
@using (Html.BeginForm("Index", "Missions", FormMethod.Get))
{
@Html.EditorFor(x => x.Decision)
}
</nav>
<div id="content" class="container">
<table class="table table-bordered">
<tr>
<th class="col-md-2">
Decision
</th>

</tr>
@foreach (var item in Model.OnePageOfMissions)
{
<tr>
<td class="col-md-1">
@Html.DisplayFor(modelItem => item.decision)
</td>
</tr>
}
@Html.PagedListPager((IPagedList)Model.OnePageOfMissions, page => Url.Action("Index", new { page, Decision = Model.Decision}))
</table>
</div>




I created a template to display checkboxlist like this:



<div class="form-check row">
@Html.HiddenFor(x => x.ID)
@Html.CheckBoxFor(x => x.IsChecked, htmlAttributes: new { onchange = "form.submit();", @class = "form-check-input col-md-2" })
@Html.LabelFor(x => x.Display, Model.Display, htmlAttributes: new { @class = "form-check-label col-md-8" })




My controller looks like this:



 public ActionResult Index(int? page, List<CheckBoxListItem> Decision)
{
IndexViewModel model = new IndexViewModel();
//Display Missions
model.missionsList = db.missions_supportmission.ToList();
//Retrieve parameters
model.Decision = Decision;
//PagedList
var pageNumber = page ?? 1; // if no page was specified in the querystring, default to the first page (1)
var onePageOfMissions = model.missionsList.ToPagedList(pageNumber, 10); // will only contain 10 products max because of the pageSize(equel to 10)
model.OnePageOfMissions = onePageOfMissions;
//Filter
if (model.Decision != null)
{
var selecteddecision = model.Decision.Where(x => x.IsChecked).Select(x => x.ID);
model.OnePageOfMissions = (from m in db.missions_supportmission
join l in db.list_decision
on m.decision equals l.decision_id
where selecteddecision.Contains(l.decision_id)
select m)
.OrderBy(a => a.id)
.ToPagedList(pageNumber, 10);
}
}
}
//Display CheckBox
//Checkboxlist (!important => mettre ce bloc de code après la requête link qui permet de filtrer sur les checkbox)
var allDecisions = db.list_decision.ToList();//returns List<list_decision>
var checkBoxListItems = new List<CheckBoxListItem>(); //nouvelle instance de la classe checkboxlist
model.Decision = checkBoxListItems;
foreach (var decison in allDecisions)
{//On assigne les valeurs "id", "display" et "is checked" à la variable checkboxlistitem
checkBoxListItems.Add(new CheckBoxListItem()
{
ID = decison.decision_id,
Display = decison.name_en,
IsChecked = false //On the add view, no decision are selected by default
});
}

return View(model);
}


I have also a ViewModel:



namespace MissionsDF.Models
{
public class IndexViewModel
{

public IEnumerable<missions_supportmission> missionsList { get; set; }
public List<CheckBoxListItem> Decision { get; set; }
public IndexViewModel()
{
this.Decision = new List<CheckBoxListItem>();
}
public IPagedList<missions_supportmission> OnePageOfMissions { get; set; }

}
}






asp.net-mvc filter checkboxlist pagedlist






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 7 at 15:46









AmélieV

35




35












  • You cannot pass a collection of complex objects in a url using new { Decision = Model.Decision} (look at the url it is generating) - you need to pass each property of each item in the collection to make it bind - but that would make no sense because you would probably exceed the query string limit and throw an exception. One option would be to have a (say) string SelectedDecisions property in the view model containing a comma separated string of the selected ID's, and pass that back in the url, and split it in the POST method.
    – Stephen Muecke
    Nov 7 at 20:55










  • I don't see how to do have a property containing a comma separated string of the selected ID's. Can you illustrate your idea?
    – AmélieV
    Nov 8 at 8:59










  • Add that property to your view model (and set its value using if(Decision != null){ var selected = Decision.Where(x => x.IsChecked).Select(x => x.ID); model.SelectedDecisions = String.Join(",", selected); } else { model.SelectedDecisions = selectedDecisions; }, then add a string selectedDecisions parameter to your model and in the view use @Html.PagedListPager((IPagedList)Model.OnePageOfMissions, page => Url.Action("Index", new { page, selectedDecisions = Model.SelectedDecisions}).
    – Stephen Muecke
    Nov 8 at 9:10










  • If you post the form, Decision will have a value, otherwise if you click a page number, then selectedDecisions will have the value. But your new { onchange = "form.submit(); in the CheckBoxFor() is terrible practice (and means the user can only ever click one checkbox) - you should allow the user to make their selections and then click a submit button
    – Stephen Muecke
    Nov 8 at 9:13










  • You will also need to modify the if (model.Decision != null) code so that if it is null (i.e. user selected a page number), then you would use String.Split(...) to set your var selecteddecision variable
    – Stephen Muecke
    Nov 8 at 9:20


















  • You cannot pass a collection of complex objects in a url using new { Decision = Model.Decision} (look at the url it is generating) - you need to pass each property of each item in the collection to make it bind - but that would make no sense because you would probably exceed the query string limit and throw an exception. One option would be to have a (say) string SelectedDecisions property in the view model containing a comma separated string of the selected ID's, and pass that back in the url, and split it in the POST method.
    – Stephen Muecke
    Nov 7 at 20:55










  • I don't see how to do have a property containing a comma separated string of the selected ID's. Can you illustrate your idea?
    – AmélieV
    Nov 8 at 8:59










  • Add that property to your view model (and set its value using if(Decision != null){ var selected = Decision.Where(x => x.IsChecked).Select(x => x.ID); model.SelectedDecisions = String.Join(",", selected); } else { model.SelectedDecisions = selectedDecisions; }, then add a string selectedDecisions parameter to your model and in the view use @Html.PagedListPager((IPagedList)Model.OnePageOfMissions, page => Url.Action("Index", new { page, selectedDecisions = Model.SelectedDecisions}).
    – Stephen Muecke
    Nov 8 at 9:10










  • If you post the form, Decision will have a value, otherwise if you click a page number, then selectedDecisions will have the value. But your new { onchange = "form.submit(); in the CheckBoxFor() is terrible practice (and means the user can only ever click one checkbox) - you should allow the user to make their selections and then click a submit button
    – Stephen Muecke
    Nov 8 at 9:13










  • You will also need to modify the if (model.Decision != null) code so that if it is null (i.e. user selected a page number), then you would use String.Split(...) to set your var selecteddecision variable
    – Stephen Muecke
    Nov 8 at 9:20
















You cannot pass a collection of complex objects in a url using new { Decision = Model.Decision} (look at the url it is generating) - you need to pass each property of each item in the collection to make it bind - but that would make no sense because you would probably exceed the query string limit and throw an exception. One option would be to have a (say) string SelectedDecisions property in the view model containing a comma separated string of the selected ID's, and pass that back in the url, and split it in the POST method.
– Stephen Muecke
Nov 7 at 20:55




You cannot pass a collection of complex objects in a url using new { Decision = Model.Decision} (look at the url it is generating) - you need to pass each property of each item in the collection to make it bind - but that would make no sense because you would probably exceed the query string limit and throw an exception. One option would be to have a (say) string SelectedDecisions property in the view model containing a comma separated string of the selected ID's, and pass that back in the url, and split it in the POST method.
– Stephen Muecke
Nov 7 at 20:55












I don't see how to do have a property containing a comma separated string of the selected ID's. Can you illustrate your idea?
– AmélieV
Nov 8 at 8:59




I don't see how to do have a property containing a comma separated string of the selected ID's. Can you illustrate your idea?
– AmélieV
Nov 8 at 8:59












Add that property to your view model (and set its value using if(Decision != null){ var selected = Decision.Where(x => x.IsChecked).Select(x => x.ID); model.SelectedDecisions = String.Join(",", selected); } else { model.SelectedDecisions = selectedDecisions; }, then add a string selectedDecisions parameter to your model and in the view use @Html.PagedListPager((IPagedList)Model.OnePageOfMissions, page => Url.Action("Index", new { page, selectedDecisions = Model.SelectedDecisions}).
– Stephen Muecke
Nov 8 at 9:10




Add that property to your view model (and set its value using if(Decision != null){ var selected = Decision.Where(x => x.IsChecked).Select(x => x.ID); model.SelectedDecisions = String.Join(",", selected); } else { model.SelectedDecisions = selectedDecisions; }, then add a string selectedDecisions parameter to your model and in the view use @Html.PagedListPager((IPagedList)Model.OnePageOfMissions, page => Url.Action("Index", new { page, selectedDecisions = Model.SelectedDecisions}).
– Stephen Muecke
Nov 8 at 9:10












If you post the form, Decision will have a value, otherwise if you click a page number, then selectedDecisions will have the value. But your new { onchange = "form.submit(); in the CheckBoxFor() is terrible practice (and means the user can only ever click one checkbox) - you should allow the user to make their selections and then click a submit button
– Stephen Muecke
Nov 8 at 9:13




If you post the form, Decision will have a value, otherwise if you click a page number, then selectedDecisions will have the value. But your new { onchange = "form.submit(); in the CheckBoxFor() is terrible practice (and means the user can only ever click one checkbox) - you should allow the user to make their selections and then click a submit button
– Stephen Muecke
Nov 8 at 9:13












You will also need to modify the if (model.Decision != null) code so that if it is null (i.e. user selected a page number), then you would use String.Split(...) to set your var selecteddecision variable
– Stephen Muecke
Nov 8 at 9:20




You will also need to modify the if (model.Decision != null) code so that if it is null (i.e. user selected a page number), then you would use String.Split(...) to set your var selecteddecision variable
– Stephen Muecke
Nov 8 at 9:20

















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