How to return View with JSON result

Question

Is there any way to return the View("controller", model) with JSON result? I've done like this(see code below) but it returns me an error.

    if (thereserror == true)
    {
        return Json(new
        {
            view = RenderRazorViewToString(ControllerContext, "Index", model),
            isValid = false,
            description = "Error!",
            JsonRequestBehavior.AllowGet
        });
    }
    else
    {
        return Json(new
        {
            view = RenderRazorViewToString(ControllerContext, "Index", model),
            isValid = true,
            description = "Hey!",
            JsonRequestBehavior.AllowGet
        });
    }

    private static string RenderRazorViewToString(ControllerContext controllerContext, string viewName, object model)
    {
        controllerContext.Controller.ViewData.Model = model;
        using (var sw = new StringWriter())
        {
            var ViewResult = ViewEngines.Engines.FindPartialView(controllerContext, viewName);
            var ViewContext = new ViewContext(controllerContext, ViewResult.View, controllerContext.Controller.ViewData, controllerContext.Controller.TempData, sw);
            ViewResult.View.Render(ViewContext, sw);
            ViewResult.ViewEngine.ReleaseView(controllerContext, ViewResult.View);
            return sw.GetStringBuilder().ToString();
        }
    }

For my AJAX I'm doing like this:

 $.ajax({
        type: "GET",
        url: "/serviceentry/getservice",
        data: ({ "SONumber": soNumber }),
        success: function (data) {
            if (data.isValid) {
                //I don't know what to put here
            };
        },
        error: function () {
            alert('error');
        }
    });

I saw something like this but I don't know what to do: MVC Return Partial View as JSON

You just need to redirect to specific view after you get data from ajax is this you want ? — Saineshwar Bageri - MVP, Nov 30 '17 at 06:44
What are you wanting to do with the partial view? To include it in the DOM you might do say `$(someElement).html(data.view);` — , Nov 30 '17 at 06:49
Note you have the `JsonRequestBehavior.AllowGet` in the wrong place - its `return Json(new { view = ..., description = "Hey!" }, JsonRequestBehavior.AllowGet);` — , Nov 30 '17 at 06:52
@StephenMuecke, thanks for correcting me. Actually that's one of my error. Allow Get using JsonRequestBehavior. But my concern is that, in my controller I can retrieved data from my dataset and put it in my **Model**. I want it to return just like this behavior **return View("Index", model)** — Zach, Nov 30 '17 at 07:00
So what problem are you having? And you still have not stated what your want to do with the partial view you returned — , Nov 30 '17 at 07:02
@StephenMuecke I just want to return View and data (json result) at the same time. Just like this one [link](https://www.wiliam.com.au/wiliam-blog/mvc-jsonresult-returning-view-and-data-at-the-same-time) — Zach, Nov 30 '17 at 07:08
@StephenMuecke. I change the **if** logic. I think I got it now. I just want to return JSON result with "error" description. What I did was. `if (!isError) { return View("Index", model); } else { return Json(new { view = RenderRazorViewToString(ControllerContext, "Index", model), isValid = false, description = errResult }, JsonRequestBehavior.AllowGet); }` — Zach, Nov 30 '17 at 07:24

score 1 · Accepted Answer · answered Nov 30 '17 at 07:30

JSON you are returning contains 4 properties, the way you are accessing isValid, similarly access the View.

$.ajax({
        type: "GET",
        url: "/serviceentry/getservice",
        data: ({ "SONumber": soNumber }),
        success: function (data) {
            if (data.isValid) {
                //Element- Where you want to show the partialView
                $(Element).html(data.view)
            };
        },
        error: function () {
            alert('error');
        }
    });

PS: Also Pointing the wrong placement of JSONRequestBehavior.

if (thereserror == true)
    {
        return Json(new
        {
            view = RenderRazorViewToString(ControllerContext, "Index", model),
            isValid = false,
            description = "Error!"
        },JsonRequestBehavior.AllowGet);
    }
    else
    {
        return Json(new
        {
            view = RenderRazorViewToString(ControllerContext, "Index", model),
            isValid = true,
            description = "Hey!"
        },JsonRequestBehavior.AllowGet);
    }

    private static string RenderRazorViewToString(ControllerContext controllerContext, string viewName, object model)
    {
        controllerContext.Controller.ViewData.Model = model;
        using (var sw = new StringWriter())
        {
            var ViewResult = ViewEngines.Engines.FindPartialView(controllerContext, viewName);
            var ViewContext = new ViewContext(controllerContext, ViewResult.View, controllerContext.Controller.ViewData, controllerContext.Controller.TempData, sw);
            ViewResult.View.Render(ViewContext, sw);
            ViewResult.ViewEngine.ReleaseView(controllerContext, ViewResult.View);
            return sw.GetStringBuilder().ToString();
        }
    }

Is it possible to return the whole page with model value not just the element? — Zach, Nov 30 '17 at 07:36
Since you are using AJAX here, its better that you return JSON data. You just have toinclude html(view) as one of the property in your json data. — Kumar_Vikas, Nov 30 '17 at 08:57
@Zach Yes, It's possible. You can return your html with model. Create a partial view and return the partial view with model instead of json result. — Jibin Balachandran, Nov 30 '17 at 10:29
@JibinBalachandran OP needs other properties like `isValid` , `description`(used in ajax success part). So returning just the html wont serve his cause. — Kumar_Vikas, Nov 30 '17 at 13:43

How to return View with JSON result

1 Answers1