jqGrid: how to pass complex javascript object as postData to ASP MVC controller method

Question

I need to pass a JS object to my controller method jqGrid. The object is call "activeFilters" - here is the object represented as JSON:

{"family":
     [{
      "filterDisplayName":"Performance Status",
      "filterDbName":"CurrentStatus",
      "filterValueList":"On Plan"
      }]
 }

Having problems passing the above to my jqGrid (details further down). But I can pass the object to a controller with Ajax very simply:

$.ajax({
    url: myMethodPath,
    type: 'POST',
    dataType: 'html',
    data: JSON.stringify(activeFilters),
    contentType: 'application/json; charset=utf-8',
    success: function (result) {
        alert("success")
    },
    error: function () {
        alert("Error:");
    }
});

My test controller method looks like:

    [HttpPost]
    public ActionResult DataTest(JsonFilterFamily activeFilters)
    {
        return PartialView();
    }

Add the structure of JsonFilterFamily looks like:

public class JsonFilterFamily
{
    public List<FilterFamilyMember> family { get; set; }
} 

public class FilterFamilyMember
{
    public string filterDisplayName { get; set; }
    public string filterDbName { get; set; }
    public string filterValueList { get; set; }
}

With the above Ajax, the JS object gets sent to the controller without problems. But I just can't figure out how to send the same JS object as the postData in a call to jqGrid controller. I must have the syntax wrong for post data. Here is what I am using:

$("#myJqGrid").setGridParam({ postData: { family: activeFilters.family} }).trigger("reloadGrid");

But when the controller fires, something strange happens. The debugger tells me the count of family = 1; but filterDisplayName, filterDbName, filterValueList are null. Any ideas on what is wrong?

As a solution I am just going to pass the json string to postData and deserialize in the controller method. Couple of extra steps, but it works. — rgwozdz
– rgwozdz, Commented May 8, 2012 at 17:58

Luís Cruz · Accepted Answer · 2014-12-14 13:17:13Z

1

In your controller method, try putting:

var filterFamilyMember = new FilterFamilyMember();
TryUpdateModel(filterFamilyMember);

TryUpdateModel is a MVC controller method and this will put the data in your object.

edited Dec 14, 2014 at 13:17

Luís Cruz

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answered Jun 26, 2013 at 11:59

Soumen Banerjee

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1 Comment

rgwozdz Over a year ago

Thanks for your reply - However, I don't have access to the codebase anymore, so I cannot test your suggestion!

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