VBA 字符串到日期转换，并从 now() 计算差异答案

【问题标题】：VBA String to Date Conversion, and calculate diff from now()VBA 字符串到日期转换，并从 now() 计算差异
【发布时间】：2021-07-02 10:03:56
【问题描述】：

我正在尝试以下非常简单的 VBA 代码来转换为日期时间

Sub datetimedifffference()
Dim d As String
Dim sd As Date
d = "2021-04-06T12:56:16+0000"

sd = Format(CDate(d), "mm-dd-yyyy hh:mm:ss")
Debug.Print sd
End Sub

但是它给出了类型不匹配的错误任何帮助将不胜感激

我在下面再次尝试，但时间差是减少而不是增加

Sub datetimedifffference()
Dim d As String
Dim sd As Long
d = "2021-04-06T12:56:16+0000"


sd = DateDiff("n", Now, Format(convertStringtoDate(d))

Debug.Print sd
End Sub


Function convertStringtoDate(stringdate As String) As String
    Dim strings() As String
    strings = Split(stringdate, "T")
 
    convertStringtoDate = strings(0) & " " & Left(strings(1), 8)
    'Debug.Print convertStringtoDate
End Function

【问题讨论】：

试试 IsDate 看看 d 是否可以转换成日期。

标签： vba date time timestamp timestampdiff

【解决方案1】：

T 和 +0000 应该被替换。另请注意，日期是数字，格式化日期是字符串。

Sub datetimedifffference()
    Dim d As String
    Dim sd As Date
    
    d = "2021-04-06T12:56:16+0000"
    d = Replace(d, "T", " ")
    d = Replace(d, "+0000", "")

    sd = CDate(d)
    Debug.Print sd
End Sub

【讨论】：

我尝试了一些东西。你能帮我吗
你在哪个时区？另外，请注意 convertStringtoDate 是一个非常容易误导返回 String 的函数的名称。
我在 IST +0530

【解决方案2】：

查看您的编辑，我认为您需要更改两个日期的顺序：

Sub datetimedifffference()
  Dim d As String
  Dim sd As Long
  d = "2021-04-06T12:56:16+0000"
  sd = DateDiff("n", convertStringtoDate(d), Now)
  Debug.Print sd
End Sub

Function convertStringtoDate(stringdate As String) As String
  Dim strings() As String
  strings = Split(stringdate, "T")
  convertStringtoDate = CDate(strings(0) & " " & Left(strings(1), 8))
End Function

【讨论】：

我已经更新了答案，请看一下。
如果答案有助于解决问题，请检查答案旁边的 ✓ 符号。

【解决方案3】：

你可以使用这个通用函数：

' Converts an ISO 8601 formatted date/time string to a date value.
'
' A timezone info is ignored.
' Optionally, a millisecond part can be ignored.
'
' Examples:
'   2029-02-17T19:43:08 +01.00  -> 2029-02-17 19:43:08
'   2029-02-17T19:43:08         -> 2029-02-17 19:43:08
'   ' IgnoreMilliseconds = False
'   2029-02-17T19:43:08.566     -> 2029-02-17 19:43:08.566
'   ' IgnoreMilliseconds = True
'   2029-02-17T19:43:08.566     -> 2029-02-17 19:43:08.000
'
' 2017-05-24. Gustav Brock. Cactus Data ApS, CPH.
'
Public Function CDateIso8601( _
    ByVal Expression As String, _
    Optional ByVal IgnoreMilliseconds As Boolean) _
    As Date

    Const Iso8601Separator  As String = "T"
    Const NeutralSeparator  As String = " "

    ' Length of ISO 8601 date/time string like: 2029-02-17T19:43:08 [+00.00]
    Const Iso8601Length     As Integer = 19
    ' Length of ISO 8601 date/time string like: 2029-02-17T19:43:08.566
    Const Iso8601MsecLength As Integer = 23
    
    Dim Value       As String
    Dim Result      As Date
    
    Value = Replace(Expression, Iso8601Separator, NeutralSeparator)
    If InStr(Expression, MillisecondSeparator) <> Iso8601Length + 1 Then
        IgnoreMilliseconds = True
    End If
    
    If IgnoreMilliseconds = False Then
        Result = CDateMsec(Left(Value, Iso8601MsecLength))
    Else
        Result = CDate(Left(Value, Iso8601Length))
    End If
    
    CDateIso8601 = Result
    
End Function

然后将你的格式应用到返回值。

【讨论】：