Linq，Lambda：獲取本週的所有記錄

Question

如何編寫Linq/lambda表達式來選擇本週的所有記錄。

1周=週日至週六（含）。

例如，如果今天是星期三，我應該得到本週的所有記錄，即從星期天到星期三。

那麼，我沒有寫完代碼，也沒有測試。我正在寫它，只是卡在「thisWeek」。

[HttpPost] 
public ActionResult Purchase(PurchaseSearchVM vm) 
{ 
    var filter = new PurchaseFilterRepository(); 
    var model = filter.FilterPurchase(vm); 
    return View(model); 
} 

public IQueryable<Purchase> FilterPurchase(PurchaseSearchVM vm) 
{ 
    var result = db.Purchases.AsQueryable(); 

    if (vm != null) 
    { 
     if (!string.IsNullOrEmpty(vm.option)) 
     { 
      if (vm.option == "today") 
      { 
       result = result.Where(p => p.Date.Date == DateTime.Now.Date); 
      } 
      else if (vm.option == "yesterday") 
      { 
       result = result.Where(p => p.Date.Date == DateTime.Now.Date.AddDays(-1)); 
      }      
      else if (vm.option == "thisWeek") 
      { 
       //help needed here 
       result = result.Where(p=> p.Date.) 
      } 
      else if (vm.option == "thisMonth") { } 
      else if (vm.option == "lastMonth") { } 
      else if (vm.option == "thisYear") { } 
      else if (vm.option == "lastYear") { } 
     } 

     if (!string.IsNullOrEmpty(vm.supplier)) 
     { 
      var value = vm.supplier; 
      //query here 
     } 

     if (vm.fromDate != null || vm.toDate != null) 
     { 
      if (vm.fromDate != null && vm.toDate == null) 
      { 
       //query here 
      } 
      else if (vm.fromDate == null && vm.toDate != null) 
      { 
       //query here 
      } 
      else if (vm.fromDate != null && vm.toDate != null) 
      { 
       //query here 
      } 
     } 

     if (vm.IsPaid != null) 
     { 
      //query here 
     } 
    } 
    return result; 
} 

UI

Answer 1

可以計算startDayOfWeek和endDayOfWeek這樣的： -

DateTime startDayOfWeek= DateTime.Today.AddDays(-1 * (int)(DateTime.Today.DayOfWeek)); 
DateTime endDayOfWeek = DateTime.Today.AddDays(6 - (int)DateTime.Today.DayOfWeek); 

然後簡單地套用使用Where過濾器： -

var result = vm.Where(x => x.myDate >= startDayOfWeek && x.myDate <= endDayOfWeek); 

Answer 2

請用下面的代碼片段嘗試一下。

namespace ConsoleApplication1 
{ 
    class Program 
    { 
     static void Main(string[] args) 
     { 
      List<Purchase> dates = new List<Purchase>(); 
      dates.Add(new Purchase() { FromDate = DateTime.Now.AddDays(-6) }); 
      dates.Add(new Purchase() { FromDate = DateTime.Now.AddDays(-5) }); 
      dates.Add(new Purchase() { FromDate = DateTime.Now.AddDays(-4) }); 
      dates.Add(new Purchase() { FromDate = DateTime.Now.AddDays(-3) }); 
      dates.Add(new Purchase() { FromDate = DateTime.Now.AddDays(-2) }); 
      dates.Add(new Purchase() { FromDate = DateTime.Now.AddDays(-1) }); 
      dates.Add(new Purchase() { FromDate = DateTime.Now.AddDays(0) }); 
      dates.Add(new Purchase() { FromDate = DateTime.Now.AddDays(1) }); 
      dates.Add(new Purchase() { FromDate = DateTime.Now.AddDays(2) }); 
      dates.Add(new Purchase() { FromDate = DateTime.Now.AddDays(7) }); 
      dates.Add(new Purchase() { FromDate = DateTime.Now.AddDays(8) }); 

      var test = dates.Where(i => i.FromDate >= DateTime.Now.StartOfWeek(DayOfWeek.Sunday) && i.FromDate <= DateTime.Now.StartOfWeek(DayOfWeek.Sunday).AddDays(7)).ToList(); 
      // you can access the this weeks dates here 
      Console.ReadLine(); 
     } 
    } 

    public class Purchase 
    { 
     public DateTime FromDate { get; set; } 
    } 
    public static class DateTimeExtensions 
    { 
     public static DateTime StartOfWeek(this DateTime dt, DayOfWeek startOfWeek) 
     { 
      int diff = dt.DayOfWeek - startOfWeek; 
      if (diff < 0) 
      { 
       diff += 7; 
      } 

      return dt.AddDays(-1 * diff).Date; 
     } 
    } 
} 

Answer 3

可以按如下步驟來去除天的大寫金額，直到本週初寫：

result = result.Where(
       p=> p.Date >= p.Date.AddDays(-(p.Date.DayOfWeek - DayOfWeek.Sunday) && 
       p.Date <= DateTime.Today.Date); 

，或者使其更具全球：

var firstDayOfWeek = CultureInfo.CurrentCulture.DateTimeFormat.FirstDayOfWeek; 
result = result.Where(
       p=> p.Date >= p.Date.AddDays(-(p.Date.DayOfWeek - firstDayOfWeek) && 
       p.Date <= DateTime.Today.Date);