使用LINQ的分层数据表示
本文关键字:数据表示 分层 LINQ 使用 | 更新日期: 2023-09-27 18:02:50
总之,我正试图用LINQ对下面的类进行分层数据表示。有谁能帮我一下吗
public class Employee
{
public Employee(int empId,int? managerId)
{
this.Id = empId;
this.MangerId = managerId;
this.Children = new List<Employee>();
}
public int Id { get; set; }
public int? MangerId { get; set; }
public string EmployeeName { get; set; }
public List<Employee> Children { get; set; }
}
示例数据var empList = new List<Employee>
{
new Employee(1,2){EmployeeName = "Joseph"},
new Employee(2,3){EmployeeName = "Smith"},
new Employee(3,4){EmployeeName = "Bob"},
new Employee(4,null){EmployeeName = "Doug"},
new Employee(5,2){EmployeeName = "Dave"},
new Employee(6,4){EmployeeName = "Allan"}
};
,输出应该像这样
/*道格鲍勃艾伦史密斯约瑟夫戴夫
*/
如有任何帮助,不胜感激
edit:高层员工的managerId将为null
使用您的模型,Employee模型不会直接链接到它们自己。相反,我们必须使用它们的标识符遍历层次结构。
首先,我们得到根雇员:var rootEmp = EmpList.Single(e => e.MangerId == null);
然后,使用递归函数遍历层次结构:
string WalkEmployees(Employee root)
{
// Create the container of the names
var builder = new StringBuilder();
// Get the children of this employee
var children = EmpList.Where(e => e.MangerId == root.Id);
// Add the name of the current employee in the container
builder.Append(root.EmployeeName + " ");
// For each children, walk them recursively
foreach (var employee in children)
{
builder.Append(WalkEmployees(employee));
}
// Return the container of names
return builder.ToString();
}
最后,调用函数:
WalkEmployees(rootEmp);
本质上,递归函数垂直遍历层次结构:
Doug
——鲍勃
——史密斯
——约瑟夫
——Dave
——艾伦
尽管如此,为了让Allan紧跟在Bob之后,您希望进行水平行走。为此,我为您的员工添加了一个视图模型,该模型描述了他们在层次结构中的级别。
public class EmployeeViewModel
{
public EmployeeViewModel(Employee employee, int level)
{
Employee = employee;
Level = level;
}
public Employee Employee { get; set; }
public int Level { get; set; }
}
员工步行功能变为:
IEnumerable<EmployeeViewModel> WalkEmployees(Employee root, int level)
{
// Create the container of the employees
var container = new List<EmployeeViewModel> {new EmployeeViewModel(root, level)};
// Get the children of this employee
var children = EmpList.Where(e => e.MangerId == root.Id);
// For each children, walk them recursively
foreach (var employee in children)
{
container.AddRange(WalkEmployees(employee, level + 1));
}
// Return the container
return container;
}
及其调用:
var rootEmp = EmpList.Single(e => e.MangerId == null);
var employees = WalkEmployees(rootEmp, 0);
// Order the employees by its level in the hierarchy
var orderedEmployees = employees.OrderBy(vm => vm.Level);
// Display the names
foreach (var orderedEmployee in orderedEmployees)
{
Console.Write(orderedEmployee.Employee.EmployeeName + " ");
}
你会得到这样的结果:
Doug
——鲍勃
——艾伦
——史密斯
——约瑟夫
——Dave
你们的模型很难处理,因为模型之间缺乏联系。这里有一个更有力的建议:
public class Employee
{
#region Constructors
public Employee()
{
Employees = new List<Employee>();
}
public Employee(string name) : this()
{
Name = name;
}
public Employee(string name, Employee manager) : this(name)
{
Manager = manager;
}
public Employee(string name, Employee manager, params Employee[] employees) : this(name, manager)
{
Employees.AddRange(employees);
}
#endregion
#region Properties
public List<Employee> Employees { get; set; }
public int Id { get; set; }
public Employee Manager { get; set; }
public string Name { get; set; }
#endregion
}
您现在可以像这样生成您的员工:
/// <summary>
/// Generates the employees in a hierarchy way.
/// </summary>
/// <returns>Returns the root employee.</returns>
Employee GenerateEmployees()
{
var doug = new Employee("Doug");
doug.Employees.Add(new Employee("Allan", doug));
var bob = new Employee("Bob", doug);
doug.Employees.Add(bob);
var smith = new Employee("Smith", bob);
bob.Employees.Add(smith);
smith.Employees.Add(new Employee("Joseph", smith));
smith.Employees.Add(new Employee("Dave", smith));
return doug;
}
walk函数变为:
string WalkEmployees(Employee root)
{
var builder = new StringBuilder();
builder.Append(root.Name + " ");
foreach (var employee in root.Employees)
{
builder.Append(WalkEmployees(employee));
}
return builder.ToString();
}
如果你使用EntityFramework来设计一个使用导航属性的数据库,这个实现会更有意义。
尝试:
string text = string.Join(" ",
from i in empList
orderby i.MangerId.HasValue, i.MangerId descending
select i.EmployeeName);