c# - MVCjqDrid store GridSettings in session -


i have mvcjqgrid:

@(html.grid("datagrid")           .setjsonreader(new mvcjqgrid.datareaders.jsonreader { id = "id", repeatitems = false })           .setrequesttype(requesttype.post)           .addcolumn(new column("name").setlabel("name").setsearch(true))           .addcolumn(new column("email").setlabel("e-mail").setsearch(true).setformatter(formatters.email))           .addcolumn(new column("phone").setlabel("phone").setsearch(true))           .setsearchtoolbar(true)           .seturl(url.action("getdata", "controller"))           .setsearchonenter(false)           .setrownum(10)           .setrowlist(new[] { 10, 15, 20, 50 })           .setviewrecords(true)           .setpager("pager"))

and controller:

    public actionresult getdata()     {         return view(new myentity[0]);     }      [httppost]     public jsonresult getdata(gridsettings gridsettings)     {         int totalrecords = datahelper.getcount();         var data = datahelper.getdata(gridsettings);         var jsondata = new         {             total = totalrecords / gridsettings.pagesize + 1,             page = gridsettings.pageindex,             records = totalrecords,             rows = data         };         return json(jsondata);     }

edit:
question how can store gridsettings in session in right way, need every time user backs page, page should same when left?
if do:

session["settings"] = gridsettings;

i need way compare stored gridsettings 1 passed action.

why don't use http cache case? can write caching provider, , http cache 1 implementation this. in future can extend more providers it.


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