Question

有没有办法用JSON结果返回View（“controller”，model）？我这样做了（见下面的代码），但它给我一个错误。

    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();
        }
    }

对于我的AJAX，我这样做：

 $.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');
        }
    });

我看到类似的东西，但我不知道该怎么做：MVC Return Partial View as JSON

Answer 1

您返回的

JSON包含4个属性，即您访问isValid的方式，同样可以访问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：同时指出JSONRequestBehavior的错误位置。

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();
        }
    }

如何使用JSON结果返回View

1 个答案: