Question

我一直在寻找各种主题，但没有找到答案所以我问：是否有可能在VB中检测到白天之间的差异？我的意思是，我在星期一运行程序，今天他会告诉我星期一，是天差（昨天是星期天），变量将被重置，因此每天都要变量有效一天。

Answer 1

你需要这样的东西（不记得你是否可以在vb.net中声明今天这样的日期，但其余的代码应该是合理的）：

dim tDay as datetime = now()
Select Case tDay.DayOfWeek
Case DayOfWeek.Monday
'Its Monday
Case DayOfWeek.Tuesday
'Its Tuesday
.
.
.
.
End Select

未经测试，但会让你找到合适的路线

编辑：要检查日期是否已更改并重新分配变量，请将日期和x的变量声明为系统设置（My.Settings）。然后，从那里增加上述情况：

Case DayOfWeek.Monday
If My.Settings.weekDayCheck <> DayOfWeek.Monday Then
My.Settings.weekDayCheck = DayOfWeek.Monday
My.Settings.x = 20
Else
My.Settings.x = My.Settings.x + 1
End If

等等

Answer 2

棘手的问题。听起来你需要在执行之间存储程序的状态。您可以通过将数据保存到磁盘来执行此操作，例如通过使用序列化（http://en.wikipedia.org/wiki/Serialization），然后在下次加载程序时反序列化它（如果它存在于磁盘上）。

从那时起，这只是检查昨天的日期：

Sub Main
    Dim storedDate As DateTime = DateTime.Now.AddDays(-1)
    Console.WriteLine(DateTime.Now.Date.Subtract(storedDate.Date).TotalDays)
End Sub

以下是一个例子：

Sub Main
    Dim state As new State()

    StateHelper.SaveState("C:\\Users\\a-h\\test.xml", state)
    state = StateHelper.LoadState("C:\\Users\\a-h\\test.xml")

    Console.WriteLine(DateTime.Now.Date.Subtract(state.DateLastAccessed.Date).TotalDays)
End Sub

Class StateHelper
    Shared Sub SaveState (ByVal fileName As String, ByVal state As State)
        Dim serializer = new XmlSerializer(GetType(State))  

        Using fs As New FileStream(fileName, FileMode.Create)
            serializer.Serialize(fs, state)
        End Using
    End Sub

    Shared Function LoadState (ByVal fileName As String) As State
        If File.Exists(fileName)
            Dim serializer = new XmlSerializer(GetType(State))  

            Using fs As New FileStream(fileName, FileMode.Open)
                Return serializer.Deserialize(fs)
            End Using
        Else
            Return Nothing
        End If
    End Function 
End Class

Class State
    Property DateLastAccessed As DateTime = DateTime.Now
    Property Name As String = Environment.UserName
End Class

VB检测天数之间的差异

2 个答案: