停止准时事件

Question

经过大量的文章搜索后，我仍然无法得到我的头。请告诉我如何停止Application.Ontime事件。我遇到过解释To cancel a pending OnTime event, you must provide the exact time that it is scheduled to run

现在我的问题是我应该提供第一次运行事件的时间，还是应该提供下一次触发事件的时间？

我在下面的代码都尝试版本StopTimer的他们都让我"Method OnTime of object _Application failed

Option Explicit 

Private Sub Workbook_Open() 
    count = 1 
    Call test 
End Sub 



Public runwhen As Double 
Public Const runwhat As String = "TheSub" 
Public firstrunTime As Double 
Public count As Integer 




Sub test() 
    If count = 1 Then 
     runwhen = Now + TimeSerial(0, 0, 5) 
     firstrunTime = runwhen 
    Else 
     runwhen = Now + TimeSerial(0, 0, 5) 
    End If 

    Application.OnTime runwhen, "TheSub" 
End Sub 

Sub TheSub() 
    MsgBox "Hi!!!!!!" 
    count = count + 1 
    Call test 
    If count = 5 Then 
     StopTimer 
    End If 
End Sub 

'First Version of StopTimer 
Sub StopTimer() 
    Application.OnTime firstrunTime, "TheSub", , False 

End Sub 

'Second Version of StopTimer 
Sub StopTimer() 
    runwhen=now+TimeSerial(0,0,5) 
    Application.OnTime runwhen, "TheSub", , False 
End Sub 

请的是，我很想念以下工艺试验和

我已经修改填写我现在用的STOPTIMER的第三个版本，但仍然是我的代码给了我在我的机器上同样的错误

Sub test() 
    If count = 1 Then 
     runwhen = Now + TimeSerial(0, 0, 5) 
     firstrunTime = runwhen 
    Else 
     runwhen = Now + TimeSerial(0, 0, 5) 
    End If 
    If count <> 5 Then 
     Application.OnTime runwhen, "TheSub" 
    Else 
     Call StopTimer 
    End If 
End Sub 

Answer 1

要取消OnTime事件，您需要通知计划运行的时间。

您的第一次尝试是告诉它取消不再计划的预定事件 - 实际上它可能发生在几个小时前。

您的第二次尝试是告诉它取消预定的事件，该事件在您决定要取消事件后5秒内发生。你可能是幸运的，并设法决定取消它很快，你设置5秒是正确的时间，但你可能不会。 （这取决于时钟的准确度以及计算机执行代码的速度。）

您需要做的是告诉它在您设置的同一时间取消该事件，因此您的代码需要说：

'Third Version of StopTimer 
Sub StopTimer() 
    Application.OnTime runwhen, "TheSub", , False 
End Sub 

该版本将使用相同的时间（runwhen）在取消当您设置的测试子程序时使用。

---

更新重新新代码：

代码的最初版本会工作（与StopTimer的版本3），不过新版本失败，因为你已经改变了它设置runwhen你的时候不应该。

让我们逐步了解新版本代码中正在发生的事情。假设您在上午6:00:00打开工作簿，并且CPU速度非常慢，以便我们可以为各种事件分配各种时间。

06:00:00.000 - Workbook opens 
06:00:00.001 - Subroutine Test is called 
06:00:00.002 - Count is 1, so first If statement executes the first section 
06:00:00.003 - runwhen is set to 06:00:05.003 
06:00:00.004 - firstruntime is set to 06:00:05.003 
06:00:00.005 - Count is 1, not 5, so second If statement executes the first section 
06:00:00.006 - OnTime is set to run TheSub at 06:00:05.003 
06:00:00.007 - Subroutine Test finishes and control returns to TheSub 
06:00:00.008 - Count is 1, not 5, so If statement is not executed 
06:00:00.009 - Subroutine TheSub finishes and execution of macro stops 

06:00:05.003 - OnTime event triggers 
06:00:05.004 - Subroutine TheSub is called 
06:00:05.005 - MsgBox is displayed 
       The user is very slow to press the button this time. (Mainly because I had 
       written a lot of the following times, and then realised my Count was out 
       by 1, and I didn't want to have to rewrite everything - so I just added 
       a very slow response here.) 
06:00:12.000 - User presses OK 
06:00:12.001 - Count is set to 2 
06:00:12.002 - Subroutine Test is called 
06:00:12.003 - Count is 2, not 1, so first If statement falls into Else portion 
06:00:12.004 - runwhen is set to 06:00:17.004 
06:00:12.005 - Count is 2, not 5, so second If statement executes the first section 
06:00:12.006 - OnTime is set to run TheSub at 06:00:17.004 
06:00:12.007 - Subroutine Test finishes and control returns to TheSub 
06:00:12.008 - Count is 2, not 5, so If statement is not executed 
06:00:12.009 - Subroutine TheSub finishes and execution of macro stops 

06:00:17.004 - OnTime event triggers 
06:00:17.005 - Subroutine TheSub is called 
06:00:17.006 - MsgBox is displayed 
06:00:18.000 - User presses OK 
06:00:18.001 - Count is set to 3 
06:00:18.002 - Subroutine Test is called 
06:00:18.003 - Count is 3, not 1, so first If statement falls into Else portion 
06:00:18.004 - runwhen is set to 06:00:23.004 
06:00:18.005 - Count is 3, not 5, so second If statement executes the first section 
06:00:18.006 - OnTime is set to run TheSub at 06:00:23.004 
06:00:18.007 - Subroutine Test finishes and control returns to TheSub 
06:00:18.008 - Count is 3, not 5, so If statement is not executed 
06:00:18.009 - Subroutine TheSub finishes and execution of macro stops 

06:00:23.004 - OnTime event triggers 
06:00:23.005 - Subroutine TheSub is called 
06:00:23.006 - MsgBox is displayed 
06:00:24.000 - User presses OK 
06:00:24.001 - Count is set to 4 
06:00:24.002 - Subroutine Test is called 
06:00:24.003 - Count is 4, not 1, so first If statement falls into Else portion 
06:00:24.004 - runwhen is set to 06:00:29.004 
06:00:24.005 - Count is 4, not 5, so second If statement executes the first section 
06:00:24.006 - OnTime is set to run TheSub at 06:00:29.004 
06:00:24.007 - Subroutine Test finishes and control returns to TheSub 
06:00:24.008 - Count is 4, not 5, so If statement is not executed 
06:00:24.009 - Subroutine TheSub finishes and execution of macro stops 

06:00:29.004 - OnTime event triggers 
06:00:29.005 - Subroutine TheSub is called 
06:00:29.006 - MsgBox is displayed 
06:00:30.000 - User presses OK 
06:00:30.001 - Count is set to 5 
06:00:30.002 - Subroutine Test is called 
06:00:30.003 - Count is 5, not 1, so first If statement falls into Else portion 
06:00:30.004 - runwhen is set to 06:00:35.004 
06:00:30.005 - Count is 5, so second If statement executes falls into the Else portion 
06:00:30.006 - Subroutine StopTimer is called 
06:00:30.007 - Code attempts to cancel Ontime event scheduled for 06:00:35.004 (the value of runwhen), 
       but fails because no such event is scheduled) 

的失败是因为你更新的runwhen值（在06：00：在我的例子30.004），但后来不设置导通时间的事件。然后你去取消这个活动，但它不能取消。

当您设置OnTime事件时，您应该设置runwhen只有，然后您将能够使用该变量取消该事件。

我建议你改变你的整个代码为：

'In your Workbook module 
Option Explicit 

Private Sub Workbook_Open() 
    count = 1 
    Call StartTimer 
End Sub 

'In your main code module 
Option Explicit 

Public runwhen As Double 
Public count As Integer 

Sub TheSub() 
    MsgBox "Hi!!!!!!" 
    count = count + 1 
    Call StartTimer 
End Sub 

Sub StartTimer() 
    If count <> 5 Then 
     runwhen = Now + TimeSerial(0, 0, 5) 
     Application.OnTime runwhen, "TheSub" 
    End If 
End Sub 

如果设置了这种方式，你不需要StopTimer子程序，因为你只启动定时器，你希望它的次数跑。

但是，您可能正试图设计一个系统，用户可以决定何时停止计时器，也许可以通过点击某个按钮。如果是这样，你只需要在按钮的代码，以包括以下语句停止定时器：

Application.OnTime runwhen, "TheSub", , False 

Answer 2

你的代码工作正常。在这里，我使用StopTimer子例程的第二个版本，并将所有代码放入标准代码模块中。我想在你的代码的罪魁祸首是你不声明全局变量：

Public runwhen As Double 
Public Const runwhat As String = "TheSub" 
Public firstrunTime As Double 
Public count As Integer 

您的代码模块的顶部。把它们放在Option Explicit的正下方，以使它们正常工作。

FWIW，我解决您的StopTimer子程序的第一个版本，以使其正常工作：

Sub StopTimer() 
    On Error Resume Next 
    Application.OnTime runwhen, "TheSub", , False 
End Sub 

现在这两个子程序中可以互换使用。