最好创建一个查看模型传递到视图:
public class TempoViewModel
{
public int EmployeeId { get; set; }
public string FirstName { get; set; }
public string LastName { private get; set; }
public DateTime EmployeeStartDate { private get; set; }
//any other properties here that make up EmployeeInformation
public string EmployeeInformation
{
get
{
//Format your employee information in the way you intend for the view
return string.Format("Employee: {0}, {1}, {2}", this.FirstName, this.LastName, this.EmployeeStartDate);
}
}
}
然后有一个控制器创建视图模型:
public ViewResult Tempo()
{
employee = //logic to retrieve employee information
//map model to viewmodel
var viewModel = new TempoViewModel()
{
EmployeeId = employee.EmployeeID,
FirstName = employee.Fname,
LastName = employee.Lname, //or set whatever properties are required to display EmployeeInformation
EmployeeStartDate = employee.StartDate,
};
return View(viewModel);
}
然后在视图模型中显示view:
@model TempoViewModel
@{
ViewBag.Title = "Tempo";
}
<h2>Tempo</h2>
<div>
<h4>Tempo Employee Information</h4>
<br />
EmployeeID: @Model.EmployeeId @* Do you really want to display the employee id? *@
<br />
Fname: @Model.FirstName
<br />
Employee: @Model.EmployeeInformation
</div>
Update:
根据您当前的实现,您在致电时想要实现的目标@ViewBag.Employee
在视图中,是将模型写为字符串表示形式。在当前的实现中,要将模型转换为字符串,ToString() https://msdn.microsoft.com/en-us/library/system.object.tostring(v=vs.110).aspx调用模型的方法。由于您(可能)没有覆盖ToString()
方法,而是调用继承的对象实现,它写出完整的名称空间和类名(这就是我假设您说路径时的意思)。
要纠正您当前的解决方案,您可以添加一个实现ToString()
到你的 Employee 类。例如:
public class Employee
{
public Employee(int employeeId, string firstName, string lastName)
{
this.EmployeeId = employeeId;
this.FName = firstName;
this.LName = lastName;
//etc
}
public int EmployeeId { get; set; }
public string FName { get; set; }
public string LName { get; set; }
public override string ToString()
{
//Implement your required string representation of the object here
return string.Format("EmployeeId: {0}, FName: {1}, LName: {2}", EmployeeId, FName, LName);
}
}