c# - Choose One Model in List<t> from view and send back to action -

as know can pass data controller views using viewdata , viewbag, tempdata ,session , model. used model send list of dynamically created list object view.

model :

public class person {     public int id { get; set; }     public string name { get; set; }     public string family { get; set; }     public string link { get; set; } } 

controller :

list<person> lstperson = new list<person>(); lstperson.add(new person() { id = 1, name = "n1", link = "l1" }); lstperson.add(new person() { id = 2, name = "n2", link = "l2" }); lstperson.add(new person() { id = 3, name = "n3", link = "l3" }); lstperson.add(new person() { id = 4, name = "n4", link = "l4" }); return view(lstperson); 

fill dropdownlist binding model list.

@model list<mvcapplication2.models.person> <h2>index</h2> @html.beginform("send model"){  @html.dropdownlist(     "foo",     new selectlist(         model.select(x => new { value = x.id, text = x.name }),         "value",         "text"     )  ) <input type="submit" value="send" /> } 

sending list of model possible using @html.actionlink("index","index",lstperson), how know item selected? best way catch model that's selected in dropdownlist , send controller , work selected model in list?

model:

public class peopleviewmodel {    public list<selectlistitem> people {get;set;}    public int selectedpersonid {get;set;} } 

controller:

public actionresult people() {     list<person> lstperson = new list<person>();     lstperson.add(new person() { id = 1, name = "n1", link = "l1" });     lstperson.add(new person() { id = 2, name = "n2", link = "l2" });     lstperson.add(new person() { id = 3, name = "n3", link = "l3" });     lstperson.add(new person() { id = 4, name = "n4", link = "l4" });     tempdata["peoplelist"] = lstperson;     var model = new peopleviewmodel     {        people = lstperson.select(        p => new selectlistitem{ value = p.id.tostring(), text = p.name}).tolist()     }     return view(model); } 

view:

@model peopleviewmodel <h2>index</h2> @html.beginform("send model"){  @html.dropdownlistfor(m=>m.selectedpersonid,model.people)  ) <input type="submit" value="send" /> } 

controller post action

[httppost] public actionresult people(peopleviewmodel model) {     var peoplelist = (list<person>)tempdata["peoplelist"];     var selectedperson = peoplelist.firstordefault(p=>p.id == model.selectedpersonid); } 

send list of model possible using @html.actionlink("index","index",lstperson).

actually wrong assumption:

1. html.actionlink generates link : <a href="">
2. only inputs posted controller, if want post should rendered <input> or <select>