[英]Excel VBA datediff to Calculate Months, Days and Month together?
我正在尝试使用Datediff函数计算总月数,天数和小时数所用的时间。 这不可能吗?
DateDiff("d hh", datein, Now)
我能做什么?
这是不可能的,因为interval参数只能是一个字符串。
你需要做更多的工作,比如获得hours
的差异,如果超过24,则将小数分隔符之前的部分转换为天数
Sub Main()
Dim d1 As Date
d1 = "15/10/2014 08:00:03"
Dim d2 As Date
d2 = Now
Dim hrsDiff As Long
hrsDiff = DateDiff("h", d1, d2)
MsgBox IIf(hrsDiff >= 24, _
hrsDiff \ 24 & " days " & hrsDiff Mod 24 & " hours", _
hrsDiff & " hours")
End Sub
这很粗糙,准备就绪,但只是方向性的。 您可以创建用户定义的函数。 这个以字符串形式返回1:2:22:15(但您可以返回包含月,日,小时,分钟变量的自定义类实例)。 它不考虑date2在date1之前(不确定会发生什么),也不考虑date1只是部分日(假设date1是午夜)。
Function MyDateDiff(date1 As Date, date2 As Date) As String
Dim intMonths As Integer
Dim datStartOfLastMonth As Date
Dim datStartOfLastHour As Date
Dim datEndOfMonth As Date
Dim intDays As Integer
Dim intHours As Integer
Dim intMinutes As Integer
Dim strResult As String
' Strip of any time
datStartOfLastMonth = DateSerial(Year(date2), Month(date2), Day(date2))
' check the dates arent in the same month
If Not ((Month(date1) = Month(date2) And Year(date1) = Year(date2))) Then
' how many months are there
intMonths = DateDiff("m", date1, date2)
Debug.Print (intMonths)
' how many days difference are there
intDays = DateDiff("d", DateAdd("m", intMonths, date1), date2)
Debug.Print (intDays)
' how many hours difference are there
intHours = DateDiff("h", datStartOfLastMonth, date2)
Debug.Print (intHours)
' how many minutes different are there
datStartOfLastHour = datStartOfLastMonth + (DatePart("h", date2) / 24)
intMinutes = DateDiff("n", datStartOfLastHour, date2)
Debug.Print (intMinutes)
Else
' Dates are in the same month
intMonths = 0
Debug.Print (intMonths)
' how many days difference are there
intDays = DateDiff("d", date1, date2)
Debug.Print (intDays)
' how many hours difference are there
intHours = DateDiff("h", datStartOfLastMonth, date2)
Debug.Print (intHours)
' how many minutes different are there
datStartOfLastHour = datStartOfLastMonth + (DatePart("h", date2) / 24)
intMinutes = DateDiff("n", datStartOfLastHour, date2)
Debug.Print (intMinutes)
End If
strResult = intMonths & ":" & intDays & ":" & intHours & ":" & intMinutes
MyDateDiff = strResult
End Function
测试这个:
?MyDateDiff(“01-SEP-2014”,“03-Oct-2014 22:15:33”)
得到:
1:2:22:15
即1个月,2天,22分15秒。
通过将组件添加回date1来反向测试它给出:
?使用DateAdd( “n” 个,15个,使用DateAdd( “H”,22,使用DateAdd( “d”,2,使用DateAdd( “M”,1 “01-SEP-2014”))))
=“03-Oct-2014 22:15:33”
如果我们在同一个月尝试2个日期:
?MyDateDiff(“01-SEP-2014”,“03-SEP-2014 22:15:33”)
我们得到:
0:2:22:15
反向测试:
?使用DateAdd( “n” 个,15个,使用DateAdd( “H”,22,使用DateAdd( “d”,2,使用DateAdd( “M”,0, “01-SEP-2014”))))
得到:
03/09/2014 22:15:00
但是你可能想要考虑日期是错误的方式...你可能只想将date1计算为部分日期,如果它在当天晚些时候开始......正如我所说,只是一个想法。
问候
一世
这可能会给你一些想法来纠正2月或闰年的日子
Private Sub CommandButton1_Click()
DoDateA
End Sub
Sub DoDateA()
Dim D1 As Date, D2 As Date, DC As Date, DS As Date
Dim CA: CA = Array("", "yyyy", "m", "d", "h", "n", "s", "s")
Dim Va%(7), Da(7) As Date, Ci%
D1 = Now + Rnd() * 420 ' vary the * factors for range of dates
D2 = Now + Rnd() * 156
If D1 > D2 Then
[b4] = "Larger"
Else
[b4] = " smaller"
DS = D1
D1 = D2
D2 = DS
End If
[d4] = D1
[e4] = D2
DC = D2
For Ci = 1 To 6
Va(Ci) = DateDiff(CA(Ci), DC, D1)
DC = DateAdd(CA(Ci), Va(Ci), DC)
Va(Ci + 1) = DateDiff(CA(Ci + 1), DC, D1)
If Va(Ci + 1) < 0 Then ' added too much
Va(Ci) = Va(Ci) - 1
DC = DateAdd(CA(Ci), -1, DC)
Cells(9, Ci + 3) = Va(Ci + 1)
Cells(8, Ci + 3) = Format(DC, "yyyy:mm:dd hh:mm:ss")
End If
Da(Ci) = DC
Cells(5, Ci + 3) = CA(Ci)
Cells(6, Ci + 3) = Va(Ci)
Cells(7, Ci + 3) = Format(Da(Ci), "yyyy:mm:dd hh:mm:ss")
Cells(10, Ci + 3) = DateDiff(CA(Ci), D2, D1)
Next Ci
End Sub
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