如何使用@Url将ViewModel从视图传递到操作。行动mvc4

I create a ViewModel:

我创建了一个视图模型:

public class TestViewModel
    {

        public int RegionId { get; set; }
        public int WeekId { get; set; }
        public int SubregionId { get; set; }
        public int YearId { get; set; }
    }

I create a Testcontroller which have two actions Index and Transfer. In the Index Action i initialize the viewModel and pass the model to index View. In the Index View I use @Url.Action("Tranfer",Model) to pass model back to Transfer action.but in the Transfer action. i can't get the correct model.Can any body tell me the solution?

我创建了一个Testcontroller，它有两个动作索引和传输。在索引操作中，我初始化viewModel并将模型传递给Index视图。在索引视图中，我使用@Url.Action(“Tranfer”，Model)将模型返回到传输操作。但在转移行动中。我找不到正确的模型。谁能告诉我解决办法吗?

Thanks!!!

谢谢! ! !

1 个解决方案

#1

Url.Action works as a Url with Query String

Url。操作作为带有查询字符串的Url工作

Query strings are like dictionary : {key, value} pairs. Therefore you cannot pass your class objects in query strings.

查询字符串类似于dictionary: {key, value}对。因此，不能在查询字符串中传递类对象。

But, what you can do, is to pass this the id for your model object and then use that id to load your object on server.

但是，您可以做的是，将这个id传递给模型对象，然后使用该id在服务器上加载您的对象。

#1