vb时间事件- 比如我想在80分钟后弹出form2

来源：百度知道 编辑：UC知道 时间：2024/05/29 06:10:22
单击command1便在80分钟后弹出form2，应该怎样才能实现这一事件？望高人指点
需要说明一下：那个“80”是从 “文本框” 获得的值。也说是时间由用户从 “文本框”输入。

Public Sub delay(Secs!) '延迟Secs毫秒
Dim T1!
T1! = Timer
lp1:
DoEvents
If (Timer - T1!) < (Secs! / 1000) Then GoTo lp1
End Sub

这样用，注意是毫秒
delay(text1.text)
form2.show

可以调用 API 得到长延时的 Timer （长达 47 天）
'模块中代码
Dim lTimerId As Long
Private Declare Function SetTimer Lib "user32" (ByVal hWnd As Long, ByVal nIDEvent As Long, ByVal uElapse As Long, ByVal lpTimerFunc As Long) As Long

Private Declare Function KillTimer Lib "user32" (ByVal hWnd As Long, ByVal nIDEvent As Long) As Long

Private Sub TimerProc(ByVal lHwnd As Long, ByVal lMsg As Long, ByVal lTimerId As Long, ByVal lTime As Long)

Dim lResult As Long
lResult = StopTimer(lTime