Question

如何在不使用ViewData和ViewBag的情况下将多个对象传递给视图？

示例：

[AllowAnonymous]
public ActionResult RandomView()
{
   // Customer.
   var customers = _context.Customers.ToList();
   List<CustomerDto> customersDto = new List<CustomerDto>();

   foreach (var customer in customers)
   {
      customersDto.Add(new CustomerDto() { 
      Id = customer.Id,
      Name = customer.Name
      });
   }

   // Reports.
   var Reports= _context.Reports.ToList();
   List<ReportsDto> reportsDto= new List<ReportsDto>();

   foreach (var report in reports)
   {
      reportsDto.Add(new ReportsDto() {
      Id = report.Id,
      Name = report.Name
      });
   }

   return View(); // How to Pass Both CustomerDto and reportDto ?
}

我想将CustomerDto和reportDto都作为强类型传递给视图。有可能吗？

Answer 1

为您的视图提供模型类。在该视图中，您引用模型类。创建模型类的两个列表属性。参见下面的示例：

public class RandomViewModel
{
    public List<CustomerDto> Customers { get; set; }
    public List<ReportDto> Reports { get; set; }
}

[AllowAnonymous]
public ActionResult RandomView()
{
   // Customer.
   var customers = _context.Customers.ToList();
   List<CustomerDto> customersDto = new List<CustomerDto>();

   foreach (var customer in customers)
   {
      customersDto.Add(new CustomerDto() { 
      Id = customer.Id,
      Name = customer.Name
      });
   }

   // Reports.
   var Reports= _context.Reports.ToList();
   List<ReportsDto> reportsDto= new List<ReportsDto>();

   foreach (var report in reports)
   {
      reportsDto.Add(new ReportsDto() {
      Id = report.Id,
      Name = report.Name
      });
   }

   var randomViewModel = new RandomViewModel() {
      Customers = customersDto,
      Reports = reportsDto
   };

   return View(randomViewModel);
}

Answer 2

您应该将这两种类型都包装到某些ViewModel类中，如下所示：

public class CustomerData
{
    public List<CustomerDto> Customer { get; set; }
    public List<ReportsDto> Reports { get; set; }
}

，然后转到您的视图：

var customerData = new CustomerData() 
{
   Customer = customersDto,
   Reports = reportsDto
}
return View(customerData);

Answer 3

方法1：

创建一个同时具有属性的视图模型并将该视图模型传递给视图

public class ViewModel
{
   public List<CustomerDto> Customers {get;set;}
   public List<ReportsDto> Reports {get;set;}
}

方法2：

使用元组并传递该元组对象以查看

var tupleModel = new Tuple<List<CustomerDto>, List<ReportsDto>>(GetCustomers(), GetReports());  
return View(tupleModel);

可见

使用Tuple强力键入视图

@model Tuple <List<CustomerDto>, List <ReportsDto>>

使用Item1和Item2属性访问两个对象

@Model.Item1
@Model.Item2

如何将多个对象从控制器动作传递到MVC中的视图？

3 个答案: