Question

我想将JsonResult传递给partialView，我能够将JsonResult返回到普通视图，但不知道如何将它传递给局部视图。传递到普通视图的JsonResult是

public JsonResult Search(int id)
{
    var query = dbentity.user.Where(c => c.UserId == id);
    return Json(query,"Record Found");
}

但想知道如何将其返回到部分视图，例如

public JsonResult Search(int id)
{
   var query = dbentity.user.Where(c => c.UserId == id);
   return PartialView(query,"Record Found");
}

Answer 1

使用行动：

public ActionResult Search(int id)
{
   var query = dbentity.user.Where(c => c.UserId == id);
   return PartialView(query);
}

并在视图中将Model转换为Json对象

<script>
var model = @Html.Raw(Json.Encode(Model))
</script>

Answer 2

根据您的评论

我想将JsonResult返回到partialView，如返回Json（PartialView，查询） - user3026519 2013年11月24日10:40

我假设你想要返回包含渲染部分视图的Json结果？话虽如此，您可以使用create helper方法将视图转换为字符串，然后将其传递给Json结果。以下是一个可能的解决方案：

你的助手方法：

/// <summary>
/// Helper method to render views/partial views to strings.
/// </summary>
/// <param name="context">The controller</param>
/// <param name="viewName">The name of the view belonging to the controller</param>
/// <param name="model">The model which is to be passed to the view, if needed.</param>
/// <returns>A view/partial view rendered as a string.</returns>
public static string RenderViewToString(ControllerContext context, string viewName, object model)
{
    if (string.IsNullOrEmpty(viewName))
        viewName = context.RouteData.GetRequiredString("action");

    var viewData = new ViewDataDictionary(model);

    using (var sw = new StringWriter())
    {
        var viewResult = ViewEngines.Engines.FindPartialView(context, viewName);
        var viewContext = new ViewContext(context, viewResult.View, viewData, new TempDataDictionary(), sw);
        viewResult.View.Render(viewContext, sw);

        return sw.GetStringBuilder().ToString();
    }

致电行动：

public ActionResult Search(int id)
{
    var query = dbentity.user.Where(c => c.UserId == id);
    return Json(RenderViewToString(this.ControllerContext, "Search", query));
}

如何在asp.net mvc中将JsonResult传递给PartialView

2 个答案: