Question

我在ColumnA中都有各种日期格式的字符串形式的日期。这是一个小例子。

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Call by Name false
Time 64ms
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Call by Value false
Time 3068ms
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Answer 1

我试图用正则表达式获取日期。据我所知，日期要么在字符串的末尾，要么在字符串的后面有一个点（后跟扩展名）。正则表达式考虑了年份的长度：4或2。还可以管理月份以文本形式表示的情况。 重要！当一天缺席时，我将其设置为第一天。

Function GetDate(strString$)
    Dim sYear$, sMonth$, sDay$
    With CreateObject("VBScript.RegExp")
        .IgnoreCase = True: .Pattern = "(\d{2})([a-z]{3}|\d{2})(\d{2})?(?=\.|$)"
        With .Execute(strString)
            If .Count > 0 Then
                With .Item(0)
                    sYear = .SubMatches(0)
                    sMonth = .SubMatches(1)
                    sDay = .SubMatches(2)
                    sYear = "20" & sYear
                    If Not IsNumeric(sMonth) Then
                        sMonth = GetMonthIndex(sMonth)
                    End If
                    If Len(sDay) = 0 Then sDay = "01"
                    GetDate = DateSerial(CInt(sYear), CInt(sMonth), CInt(sDay))
                End With
            End If
        End With
    End With
End Function

Private Function GetMonthIndex$(strMonth$)
    Select Case strMonth
        Case "Jan": GetMonthIndex = "01"
        Case "Feb": GetMonthIndex = "02"
        Case "Mar": GetMonthIndex = "03"
        Case "Apr": GetMonthIndex = "04"
        Case "May": GetMonthIndex = "05"
        Case "Jun": GetMonthIndex = "06"
        Case "Jul": GetMonthIndex = "07"
        Case "Aug": GetMonthIndex = "08"
        Case "Sep": GetMonthIndex = "09"
        Case "Nov": GetMonthIndex = "10"
        Case "Oct": GetMonthIndex = "11"
        Case "Dec": GetMonthIndex = "12"
    End Select
End Function

Answer 2

尝试

Sub test()
    Dim rngDB As Range
    Dim vDB As Variant, vR() As Variant
    Dim vS As Variant
    Dim i As Long, j As Integer, c As Integer
    Dim s As String

    Set rngDB = Range("a1", Range("a" & Rows.Count).End(xlUp))
    vDB = rngDB

    r = UBound(vDB, 1)
    ReDim vR(1 To r, 1 To 1)

    For i = 1 To r
        s = vDB(i, 1)
        vS = Split(s, ".")
        If IsNumeric(vS(1)) Then
            s = vS(1)
            s = Right(s, 6)
            s = Left(s, 2) & " " & Mid(s, 3, 2) & " " & Right(s, 2)
            vR(i, 1) = DateValue(s)
        Else
            s = vS(0)
            s = Right(s, 6)
            If IsNumeric(s) Then
                s = Left(s, 2) & " " & Mid(s, 3, 2) & " " & Right(s, 2)
                vR(i, 1) = DateValue(s)
            Else
                s = Right(s, 4)
                If IsNumeric(s) Then
                    s = Left(s, 2) & " " & Mid(s, 3, 2) & " " & 1
                    vR(i, 1) = DateValue(s)
                Else
                    s = vS(0)
                    s = Right(s, 7)
                    s = Left(s, 2) & " " & Mid(s, 3, 3) & " " & Right(s, 2)
                    vR(i, 1) = DateValue(s)
                End If
            End If
        End If
    Next i
    Range("b1").Resize(r) = vR

End Sub

Answer 3

没有任何固定的模式将很复杂。在您的情况下，从逻辑上来说，更好的方法是将日期模式与

这样的文件相关联

1 => * .BUS-> YYMM
2 => * .IDE-> YYMMDD
3 => TTN *-> YYYYMMDD

在相同的逻辑上，获得一个映射表，其中文件名中日期字符串的位置

1 =>文件名的后4个字符
2 =>文件名的最后6个字符
3 =>扩展

在宏功能或开关功能中。

有没有办法解释任何字符串中的任何日期？

3 个答案: