ASP。将DataTime传递给视图a

I have a asp.net MVC4 web project it shows a list of production data for that day. I have added a datetime picker which allows the user to select a date that they want to show information for.

我有一个asp.net MVC4 web项目，它显示了当天的生产数据列表。我添加了一个datetime picker，它允许用户选择一个他们想要显示信息的日期。

The problem i am having is i am not sure how to go about passing the information back to the view from the method i have inside the controller.

我遇到的问题是，我不知道如何从控制器内部的方法将信息传回视图。

I have the date passing back to the controller. Inside the controller i am doing a LINQ statement that allows me to select only the production data for that day.

我将日期传回控制器。在控制器中，我正在执行LINQ语句，该语句允许我只选择当天的生产数据。

  [HttpPost]
    public ActionResult GetProductionDateInfo(string dp)
    {
        DateTime SelectedDate = Convert.ToDateTime(dp);
        DateTime SelectedDateDayShiftStart = SelectedDate.AddHours(7);
        DateTime SelectedDateDayShiftEnd = SelectedDate.AddHours(19);

        var ProductionData =

            from n in db.tbl_dppITHr
            where n.ProductionHour >= SelectedDateDayShiftStart
            where n.ProductionHour <= SelectedDateDayShiftEnd
            select n;


        return View();

I am looking to get the Var ProductionData passed back to the view so that display it inside a table.

我希望将Var ProductionData传回视图，以便在表中显示它。

4 个解决方案

#1

You can return ProductionData directly to your View.

可以将ProductionData直接返回到视图。

 return View(productionData)

And then in your View you could have @model IEnumerable<Type>

然后在你的视图中你可以有@model IEnumerable <类型> 。

However, a better practice would be to create a strongly typed ViewModel to hold the ProductionData and then return the following:

但是，更好的做法是创建一个强类型的ViewModel来保存ProductionData，然后返回以下内容:

 var model = new ProductionDataViewModel();
 model.Load();

 return View(model);

Where model a definition as follows:

模型定义如下:

public class ProductionDataViewModel { 

   public List<ProductionDataType> ProductionData { get; set; }
   public void Load() {
       ProductionData = from n in db.tbl_dppITHr
        where n.ProductionHour >= SelectedDateDayShiftStart
        where n.ProductionHour <= SelectedDateDayShiftEnd
        select n;
   }
}

Then in your view use the new strongly typed ViewModel:

然后在您的视图中使用新的强类型视图模型:

 @model ProductionDataViewModel

#2

Use a model, something like:

使用一个模型，比如:

public class ProductionDataModel
{
    //put your properties in here

    public List<ProductionData> Data { get; set; }
}

Then create/return it in your ActionResult:

然后在ActionResult中创建/返回:

var ProductionData =
    from n in db.tbl_dppITHr
    where n.ProductionHour >= SelectedDateDayShiftStart
    where n.ProductionHour <= SelectedDateDayShiftEnd
    select new ProductionData
    {
        //set properties here
    };

var model = new ProductionDataModel
{
    Data = ProductionData
};


return View(model);

Then in your view, set your model at the top:

然后在你的视图中，将你的模型放在顶部:

@model ProductionDataModel

#3

Your ProductionData variable should now be of type IEnumerbable<tbl_dppITHrRow>.

您的ProductionData变量现在应该是ienumable类型。

You can pass in the model from your controller using this code at the bottom of your action:

您可以使用下面的代码在您的控制器中传递模型:

return View(ProductionData);

In your view, you can make this your model type by placing the following Razor code in your view's .cshtml file:

在您的视图中，您可以将以下剃刀代码放在视图的.cshtml文件中，使其成为您的模型类型:

@model IEnumerbable<tbl_dppITHrRow>

Then, you can use your model in your view code:

然后，您可以在视图代码中使用您的模型:

@foreach(var row in Model) {
    <div>@row.Value</div>
}

#4

The problem here is that you are returning nothing to your view here return View(); this view just render view and no data will be passed to it.

这里的问题是您没有返回任何东西到您的视图返回视图();这个视图只是呈现视图，没有数据将被传递给它。

if ProductionData is getting values then

如果ProductionData获取值

return return View(ProductionData);

返回返回视图(ProductionData);

You can then use the values passed in the view.

然后可以使用在视图中传递的值。

#1