在 List<T> 中选择一周内没有休息日的记录 - C#答案

【问题标题】：Select records which has no day-off throughout the week in List<T> - C#在 List<T> 中选择一周内没有休息日的记录 - C#
【发布时间】：2016-08-02 14:16:17
【问题描述】：

我有一个 Employee 类，定义如下：

Employee
{
   public int Id { get; set; }
   public string Name { get; set; }
   public DateTime WorkDate { get; set; }
   public bool isOff { get; set; }
}

这是我的类实现和用法：

List<Employee> workers = new List<Employee>()
{
    new Employee { Id = 1, Name = "Emp 1", WorkDate = Convert.ToDateTime("4/11/2016"), IsOff = false},
    new Employee { Id = 1, Name = "Emp 1", WorkDate = Convert.ToDateTime("4/12/2016"), IsOff = false},
    new Employee { Id = 1, Name = "Emp 1", WorkDate = Convert.ToDateTime("4/13/2016"), IsOff = true},
    new Employee { Id = 1, Name = "Emp 1", WorkDate = Convert.ToDateTime("4/14/2016"), IsOff = false},
    new Employee { Id = 1, Name = "Emp 1", WorkDate = Convert.ToDateTime("4/15/2016"), IsOff = false},
    new Employee { Id = 1, Name = "Emp 1", WorkDate = Convert.ToDateTime("4/16/2016"), IsOff = false},
    new Employee { Id = 1, Name = "Emp 1", WorkDate = Convert.ToDateTime("4/17/2016"), IsOff = false},
    new Employee { Id = 2, Name = "Emp 2", WorkDate = Convert.ToDateTime("4/11/2016"), IsOff = false},
    new Employee { Id = 2, Name = "Emp 2", WorkDate = Convert.ToDateTime("4/12/2016"), IsOff = false},
    new Employee { Id = 2, Name = "Emp 2", WorkDate = Convert.ToDateTime("4/13/2016"), IsOff = false},
    new Employee { Id = 2, Name = "Emp 2", WorkDate = Convert.ToDateTime("4/14/2016"), IsOff = false},
    new Employee { Id = 2, Name = "Emp 2", WorkDate = Convert.ToDateTime("4/15/2016"), IsOff = false},
    new Employee { Id = 2, Name = "Emp 2", WorkDate = Convert.ToDateTime("4/16/2016"), IsOff = false},
    new Employee { Id = 2, Name = "Emp 2", WorkDate = Convert.ToDateTime("4/17/2016"), IsOff = false},
};

在上面的初始化中，只有 1 条记录是关闭的：

Id = 1, Name = "Emp 1" and the WorkDate = 4/13/2016

现在我怎样才能获得整个星期（从 4 月 11 日至 17 日）没有休息日的员工的 ID？这是Id = 2

LINQ 解决方案要好得多，但我不知道该怎么做。

更新

为避免混淆对象Employee，我将其更改为EmployeeSchedule

class EmployeeSchedule
{
   public int Id { get; set; }
   public string Name { get; set; }
   public DateTime WorkDate { get; set; }
   public bool isOff { get; set; }
}

这是实现

List<EmployeeSchedule> workers = new List<EmployeeSchedule>()
{
    new EmployeeSchedule{ Id = 1, Name = "Emp 1", WorkDate = Convert.ToDateTime("4/11/2016"), IsOff = false},
    new EmployeeSchedule{ Id = 1, Name = "Emp 1", WorkDate = Convert.ToDateTime("4/12/2016"), IsOff = false},
    new EmployeeSchedule{ Id = 1, Name = "Emp 1", WorkDate = Convert.ToDateTime("4/13/2016"), IsOff = true},
    new EmployeeSchedule{ Id = 1, Name = "Emp 1", WorkDate = Convert.ToDateTime("4/14/2016"), IsOff = false},
    new EmployeeSchedule{ Id = 1, Name = "Emp 1", WorkDate = Convert.ToDateTime("4/15/2016"), IsOff = false},
    new EmployeeSchedule{ Id = 1, Name = "Emp 1", WorkDate = Convert.ToDateTime("4/16/2016"), IsOff = false},
    new EmployeeSchedule{ Id = 1, Name = "Emp 1", WorkDate = Convert.ToDateTime("4/17/2016"), IsOff = false},
    new EmployeeSchedule{ Id = 2, Name = "Emp 2", WorkDate = Convert.ToDateTime("4/11/2016"), IsOff = false},
    new EmployeeSchedule{ Id = 2, Name = "Emp 2", WorkDate = Convert.ToDateTime("4/12/2016"), IsOff = false},
    new EmployeeSchedule{ Id = 2, Name = "Emp 2", WorkDate = Convert.ToDateTime("4/13/2016"), IsOff = false},
    new EmployeeSchedule{ Id = 2, Name = "Emp 2", WorkDate = Convert.ToDateTime("4/14/2016"), IsOff = false},
    new EmployeeSchedule{ Id = 2, Name = "Emp 2", WorkDate = Convert.ToDateTime("4/15/2016"), IsOff = false},
    new EmployeeSchedule{ Id = 2, Name = "Emp 2", WorkDate = Convert.ToDateTime("4/16/2016"), IsOff = false},
    new EmployeeSchedule{ Id = 2, Name = "Emp 2", WorkDate = Convert.ToDateTime("4/17/2016"), IsOff = false},
};

所选答案仍然适用。

【问题讨论】：

Lamda 表达式可以吗？
var list = workers.Where(x => x.IsOff == false) 我不知道必须调用什么扩展来过滤Id = 2
@Bharat 我需要LINQ 先生。
我已经添加了我的答案，..@stackoverflow.com/users/6143656/shift-n-tab

标签： c# linq ienumerable enumerable

【解决方案1】：

现在我如何获取没有休息日的员工的 ID 整个星期（从 4 月 11 日至 17 日）？ Id = 2

您可以使用Linq 扩展，并实现此目的。

var empid = workers.GroupBy(g=> g.Id)           
                   .Where(x=>x.All(e=>!e.IsOff))
                   .Select(x=>x.Key)

【讨论】：

在处理分组元素时，您需要将.Select(x=>x.Id) 更改为.Select(x=>x.Key)。

【解决方案2】：

也看看这些代码。

我已经定义了一个 Employee 的 Model 类来获取数据，

也许目前您需要 Id，但在未来您也可以使用其他属性。

public class Employee
    {
        public int Id { get; set; }
        public string Name { get; set; }
        public DateTime WorkDate { get; set; }
        public bool IsOff { get; set; }
    }

和带有 Linq 表达式的代码，

 Employee objEmployee = new Employee();
 objEmployee.Id = workers
         .GroupBy(itm => itm.Id)
         .Where(itm => !itm.Any(m => m.IsOff))
         .Select(itm => itm.Key)
         .FirstOrDefault();

【讨论】：

这个也很有用

【解决方案3】：

这意味着您需要一个fromDate、toDate，并且您必须从列表中找到员工，条件是，WorkDate 应该在fromDate 和toDate 之间，并且还有@ 987654326@ 表示has no day off throughout the week 所以你可以使用下面的sn-p：

 DateTime fromDate = new DateTime(2016, 04, 11);
 DateTime toDate = new DateTime(2016, 04, 17);
 var noDayOfList = workers.GroupBy(x=> x.Id)
                          .Where(x =>
                                 x.All(y=> 
                                 y.WorkDate >= fromDate && 
                                 y.WorkDate <= toDate && 
                                 !y.isOff))
                          .Select(z=>z.Key).ToList();

【讨论】：

这并不是说他们在那段时间没有休息日。它将返回未关闭的日期，而不是 ID
@Rob 只需要一个 GroupBy(g=>x.id)

【解决方案4】：

你可以这样写：

var startDate = new DateTime(2016,04,11);
var endDate = new DateTime(2016,04,17);

var ids = workers
    .Where(w => w.WorkDate >= startDate)
    .Where(w => w.WorkDate <= endDate)
    .GroupBy(w => w.Id)
    .Where(g => !g.Any(w => w.IsOff))
    .Select(g => g.Key);

基本上：

过滤您需要的行（在此时间范围内）
按员工 ID 分组
确保组中没有行标记为IsOff
选择分组键（在这种情况下，它将是员工的 ID）

但请注意，调用对象 Employee 有点令人困惑，因为它真正代表的是员工的日程安排，而不是员工本身

【讨论】：

【解决方案5】：

试试看，你有两个条件 id 和 date，但你的 id 足以满足你的要求，

List<string> resEmp = new List<string>();

foreach (var item in workers )
 {
     if (item.Id == 2 and item.IsOff == false)
      {
        resEmp.Add(item);
      }
 }

【讨论】：