[英]VBA string with milliseconds to date
I have a string in the form "yyyy-mm-dd hh:mm:ss.mmm"
(where the end is milliseconds)我有一个
"yyyy-mm-dd hh:mm:ss.mmm"
形式的字符串(结尾是毫秒)
I'd like to convert it to a number, preferably a Date
, which preserves all the information我想将它转换为一个数字,最好是一个
Date
,它保留了所有信息
I've tried CDate()
, eg.我试过
CDate()
,例如。
Dim dateValue As Date
dateValue = CDate("2017-12-23 10:29:15.223")
But get a type mismatch error但是得到一个类型不匹配的错误
A Date
type holds the number of days since December 30 1899
with a precision of one second. Date
类型保存自December 30 1899
年December 30 1899
以来的天数,精度为 1 秒。 Though it's still possible to hold the milliseconds by storing the date in a currency type since it can hold 4 extra digits compared to a Date/Double.虽然仍然可以通过将日期存储在货币类型中来保存毫秒,因为与日期/双精度相比,它可以多容纳 4 位数字。
So an alternative would be to store the date as a timestamp in a Currency
type representing the number of seconds since December 30 1899
:因此,另一种方法是将日期存储为
Currency
类型的时间戳,表示自December 30 1899
年December 30 1899
以来的秒数:
Public Function CDateEx(text As String) As Currency
Dim parts() As String
parts = Split(text, ".")
CDateEx = CCur(CDate(parts(0)) * 86400) + CCur(parts(1) / 1000)
End Function
And to convert the timestamp back to a string:并将时间戳转换回字符串:
Public Function FormatDateEx(dt As Currency) As String
FormatDateEx = Format(dt / 86400, "yyyy-mm-dd HH:mm:ss") & "." & ((dt - Fix(dt)) * 1000)
End Function
Why not use DateAdd to add the last 0.233 seconds after obtaining the whole second as a date Value?为什么不使用 DateAdd 在获得整秒作为日期值后添加最后 0.233 秒?
Dim Str As String, MS As String
Dim DateValue As Date
Dim L as Integer
Str = "2017-12-23 10:29:15.223"
For L = 1 to Len(Str)
If Left(Right(Str, L), 1) = "." Then
MS = "0" & Right(Str, L)
Str = Left(Str, Len(Str) - L)
Exit For
End If
Next L
DateValue = CDate(Str)
If MS <> "" Then DateValue = DateAdd("S",MS,DateValue)
The code below contains all the components you might need to manage your dates and their milliseconds.下面的代码包含管理日期及其毫秒可能需要的所有组件。
Private Sub ParseTime()
Dim strTime As String
Dim Sp() As String
Dim Dt As Double
strTime = "2017-12-23 10:29:15.221"
Sp = Split(strTime, ".")
strTime = Sp(0)
Dt = CDbl(CDate(strTime))
strTime = "yyyy-mm-dd hh:mm:ss"
If UBound(Sp) Then
Dt = Dt + CDbl(Sp(1)) * 1 / 24 / 60 / 60 / (10 ^ Len(Sp(1)))
strTime = strTime & "." & CInt(Sp(1))
End If
Debug.Print Format(Dt, strTime)
End Sub
I can't say that I am entirely happy with the solution because the print is only implicitly equal to the date value.我不能说我对解决方案完全满意,因为打印仅隐式等于日期值。 However, I found that the formerly valid Date/Time format, like "yyyy-mm-dd hh:mm:ss.000", doesn't seem to work since 2007. However, it should be possible to prove conclusively that the Date/Time value is equal to its rendering by the format mask I includedcd above.
但是,我发现以前有效的日期/时间格式,如“yyyy-mm-dd hh:mm:ss.000”,自 2007 年以来似乎不再有效。但是,应该可以最终证明日期/Time 值等于我上面包含的格式掩码的渲染。
Michael's answer has an error (as spotted by Jim) when the decimal part rounds up.当小数部分四舍五入时,迈克尔的回答有一个错误(由吉姆发现)。
The following corrects the error (slightly modified for tenths of seconds rather than milliseconds and with a parameterized format pattern).以下更正了错误(稍微修改了十分之一秒而不是毫秒,并使用参数化格式模式)。
Public Function FormatDateEx(dt As Currency, formatPattern As String) As String
Rem FormatDateEx = Format(dt / 86400, "yyyy-mm-dd HH:mm:ss") & "." & ((dt - Fix(dt)) * 1000)
Dim decimalPart As Double
decimalPart = Round(((dt - Fix(dt)) * 10), 0)
If (decimalPart = 10) Then
FormatDateEx = format(dt / 86400, formatPattern) & ".0"
Else
FormatDateEx = format(Fix(dt) / 86400, formatPattern) & "." & decimalPart
End If
End Function
Use the Left$
function to trim the decimal point and milliseconds:使用
Left$
函数修剪小数点和毫秒:
Dim dateValue As Date
dateValue = CDate(Left$("2017-12-23 10:29:15.223", 19))
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