CODEWITHSUNDEEP
asp.net-mvcjsonmodelcontroller
Alexander
Alexander

Reputation: 1687

MVC Controller return Content instead of JSON

I've been working on a project were all of my requirements involved JSON. However now suddenly I have a need to return results from my model that can be used in an input elements value field. I can't use the solution I have been as I get objects returned instead of plain text for the value. This is the controller pattern I have been using:

public virtual JsonResult fooData()
{
    var fooresults = new fooQueries().fooTotal();
    return new JsonResult 
        { JsonRequestBehavior = JsonRequestBehavior.AllowGet, Data = fooresults };
}

Is there a way to use return content instead of JsonResult? I'm fairly new to the .NET MVC framework and having some difficulty finding the correct way to do this.

My current results are formatted like this:

[{ "foo", 3 }]

Instead I would prefer to get plain text so that I can use an AJAX request to pass the 3 value into an input elements value="" field.

AJAX call I am using with the controller:

$.ajax({
    type: 'GET',
    url: $('#fooValue').data('url'),
    success: function (data) {
        $('#fooValue').val(data);
    }
});

The data-url is equivalent to:

../fooController/fooData

I'm just using T4MVC.

Upvotes: 0

Views: 6639

Answers (4)

Shanitoo HL
Shanitoo HL

Reputation: 1

to return a Json, the method should be changed as follow:

public JsonResult fooData()
{
    var fooresults = new fooQueries().fooTotal();
    return  Json(fooresults , JsonRequestBehavior.AllowGet);    
}

Upvotes: 0

devdigital
devdigital

Reputation: 34349

I'm not sure what the shape of fooresults is, but you should be able to alter your AJAX call to the following:

$.ajax({
   type: 'GET',
   url: $('#fooValue').data('url'),
   success: function (data) {
    $('#fooValue').val(data.foo);
   }
});

If the dataType property of the jQuery ajax call isn't explictly set, then jQuery will try and infer the type of the return result based on the MIME type, which in your case will be JSON. Therefore jQuery will deserialise the JSON to a JSON object. See http://api.jquery.com/jQuery.ajax/ for more info.

Upvotes: 0

Tiago B
Tiago B

Reputation: 2065

You can return content as:

return Content(fooresults);

But this wont be clean to you separate the elements as JSON return is.

Upvotes: 0

Brandon
Brandon

Reputation: 69953

Return a ContentResult instead of a JsonResult

public virtual ContentResult gooData()
{
    var fooresults = new fooQueries().fooTotal();
    return Content(fooresults);
}

Upvotes: 1

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