ASP.NET MVC controller actions that return JSON or

问题:

I am trying to create controller actions which will return either JSON or partial html depending upon a parameter. What is the best way to get the result returned to an MVC page asynchronously?

回答1:

In your action method, return Json(object) to return JSON to your page.

public ActionResult SomeActionMethod() {
  return Json(new {foo=\"bar\", baz=\"Blech\"});
}

Then just call the action method using Ajax. You could use one of the helper methods from the ViewPage such as

<%= Ajax.ActionLink(\"SomeActionMethod\", new AjaxOptions {OnSuccess=\"somemethod\"}) %>

SomeMethod would be a javascript method that then evaluates the Json object returned.

If you want to return a plain string, you can just use the ContentResult:

public ActionResult SomeActionMethod() {
    return Content(\"hello world!\");
}

ContentResult by default returns a text/plain as its contentType.
This is overloadable so you can also do:

return Content(\"<xml>This is poorly formatted xml.</xml>\", \"text/xml\");

回答2:

I think you should consider the AcceptTypes of the request. I am using it in my current project to return the correct content type as follows.

Your action on the controller can test it as on the request object

if (Request.AcceptTypes.Contains(\"text/html\")) {
   return View();
}
else if (Request.AcceptTypes.Contains(\"application/json\"))
{
   return Json( new { id=1, value=\"new\" } );
}
else if (Request.AcceptTypes.Contains(\"application/xml\") || 
         Request.AcceptTypes.Contains(\"text/xml\"))
{
   //
}

You can then implement the aspx of the view to cater for the partial xhtml response case.

Then in jQuery you can fetch it passing the type parameter as json:

$.get(url, null, function(data, textStatus) {
        console.log(\'got %o with status %s\', data, textStatus);
        }, \"json\"); // or xml, html, script, json, jsonp or text

Hope this helps James

回答3:

Another nice way to deal with JSON data is using the JQuery getJSON function. You can call the

public ActionResult SomeActionMethod(int id) 
{ 
    return Json(new {foo=\"bar\", baz=\"Blech\"});
}

Method from the jquery getJSON method by simply...

$.getJSON(\"../SomeActionMethod\", { id: someId },
    function(data) {
        alert(data.foo);
        alert(data.baz);
    }
);

回答4:

I found a couple of issues implementing MVC ajax GET calls with JQuery that caused me headaches so sharing solutions here.

1. Make sure to include the data type \"json\" in the ajax call. This will automatically parse the returned JSON object for you (given the server returns valid json).
2. Include the JsonRequestBehavior.AllowGet; without this MVC was returning a HTTP 500 error (with dataType: json specified on the client).
3. Add cache: false to the $.ajax call, otherwise you will ultimately get HTTP 304 responses (instead of HTTP 200 responses) and the server will not process your request.
4. Finally, the json is case sensitive, so the casing of the elements needs to match on the server side and client side.

Sample JQuery:

$.ajax({
  type: \'get\',
  dataType: \'json\',
  cache: false,
  url: \'/MyController/MyMethod\',
  data: { keyid: 1, newval: 10 },
  success: function (response, textStatus, jqXHR) {
    alert(parseInt(response.oldval) + \' changed to \' + newval);                                    
  },
  error: function(jqXHR, textStatus, errorThrown) {
    alert(\'Error - \' + errorThrown);
  }
});

Sample MVC code:

[HttpGet]
public ActionResult MyMethod(int keyid, int newval)
{
  var oldval = 0;

  using (var db = new MyContext())
  {
    var dbRecord = db.MyTable.Where(t => t.keyid == keyid).FirstOrDefault();

    if (dbRecord != null)
    {
      oldval = dbRecord.TheValue;
      dbRecord.TheValue = newval;
      db.SaveChanges();
    }
  }

    return Json(new { success = true, oldval = oldval},
                JsonRequestBehavior.AllowGet);
}

回答5:

To answer the other half of the question, you can call:

return PartialView(\"viewname\");

when you want to return partial HTML. You\'ll just have to find some way to decide whether the request wants JSON or HTML, perhaps based on a URL part/parameter.

回答6:

Alternative solution with incoding framework

Action return json

Controller

    [HttpGet]
    public ActionResult SomeActionMethod()
    {
        return IncJson(new SomeVm(){Id = 1,Name =\"Inc\"});
    }

Razor page

@using (var template = Html.Incoding().ScriptTemplate<SomeVm>(\"tmplId\"))
{
    using (var each = template.ForEach())
    {
        <span> Id: @each.For(r=>r.Id) Name: @each.For(r=>r.Name)</span>
    }
}

@(Html.When(JqueryBind.InitIncoding)
  .Do()
  .AjaxGet(Url.Action(\"SomeActionMethod\",\"SomeContoller\"))
  .OnSuccess(dsl => dsl.Self().Core()
                              .Insert
                              .WithTemplate(Selector.Jquery.Id(\"tmplId\"))
                              .Html())
  .AsHtmlAttributes()
  .ToDiv())

Action return html

Controller

    [HttpGet]
    public ActionResult SomeActionMethod()
    {
        return IncView();
    }

Razor page

@(Html.When(JqueryBind.InitIncoding)
  .Do()
  .AjaxGet(Url.Action(\"SomeActionMethod\",\"SomeContoller\"))
  .OnSuccess(dsl => dsl.Self().Core().Insert.Html())
  .AsHtmlAttributes()
  .ToDiv())

回答7:

You may want to take a look at this very helpful article which covers this very nicely!

Just thought it might help people searching for a good solution to this problem.

http://weblogs.asp.net/rashid/archive/2009/04/15/adaptive-rendering-in-asp-net-mvc.aspx

回答8:

For folks who have upgraded to MVC 3 here is a neat way Using MVC3 and Json

回答9:

PartialViewResult and JSONReuslt inherit from the base class ActionResult. so if return type is decided dynamically declare method output as ActionResult.

public ActionResult DynamicReturnType(string parameter)
        {
            if (parameter == \"JSON\")
                return Json(\"<JSON>\", JsonRequestBehavior.AllowGet);
            else if (parameter == \"PartialView\")
                return PartialView(\"<ViewName>\");
            else
                return null;


        }

回答10:

    public ActionResult GetExcelColumn()
    {            
            List<string> lstAppendColumn = new List<string>();
            lstAppendColumn.Add(\"First\");
            lstAppendColumn.Add(\"Second\");
            lstAppendColumn.Add(\"Third\");
  return Json(new { lstAppendColumn = lstAppendColumn,  Status = \"Success\" }, JsonRequestBehavior.AllowGet);
            }
        }

回答11:

Flexible approach to produce different outputs based on the request

public class AuctionsController : Controller
{
  public ActionResult Auction(long id)
  {
    var db = new DataContext();
    var auction = db.Auctions.Find(id);

    // Respond to AJAX requests
    if (Request.IsAjaxRequest())
      return PartialView(\"Auction\", auction);

    // Respond to JSON requests
    if (Request.IsJsonRequest())
      return Json(auction);

    // Default to a \"normal\" view with layout
    return View(\"Auction\", auction);
  }
}

The Request.IsAjaxRequest() method is quite simple: it merely checks the HTTP headers for the incoming request to see if the value of the X-Requested-With header is XMLHttpRequest, which is automatically appended by most browsers and AJAX frameworks.

Custom extension method to check whether the request is for json or not so that we can call it from anywhere, just like the Request.IsAjaxRequest() extension method:

using System;
using System.Web;

public static class JsonRequestExtensions
{
  public static bool IsJsonRequest(this HttpRequestBase request)
  {
    return string.Equals(request[\"format\"], \"json\");
  }
}

Source : https://www.safaribooksonline.com/library/view/programming-aspnet-mvc/9781449321932/ch06.html#_javascript_rendering