我正在阅读本应该显示的2/1/2009 - 2/28/2009的一些报告,而是显示2/1/2009 - 3/1/2009.我已经将错误缩小到这个代码,任何建议?
Function GetMonthLastDate(ByVal sDateTime As DateTime)
Try
Dim strArrMonth() As String = {"", "31", "29", "31", "30", "31", "30", "31", "31", "30", "31", "30", "31"}
Dim days As Integer
If sDateTime.Month = 2 Then
If sDateTime.Year Mod 400 = 0 Then Return sDateTime.AddDays(28)
If sDateTime.Year Mod 100 = 0 Then Return sDateTime.AddDays(27)
If sDateTime.Year Mod 4 = 0 Then Return sDateTime.AddDays(28)
End If
days = strArrMonth(sDateTime.Month)
Return Format(sDateTime.AddDays(days - 1), "MM/dd/yyyy")
Catch ex As Exception
Response.Write("<script>alert('" & ex.Message & "');</script>")
End Try
End Function
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为什么不用DateTime.DaysInMonth?这使得代码得多更加清晰.
Function GetMonthLastDate(ByVal srcDate As DateTime) As String
return _
new DateTime(srcDate.Year, srcDate.Month, _
DateTime.DaysInMonth(srcDate.Year, srcDate.Month)) _
.ToString("MM/dd/yyyy")
End Function
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您可以很容易地(也可能应该)返回DateTime本身(通过更改返回类型和nixing .ToString(...)),但我看到您的原始函数将其作为格式化字符串返回,因此为了兼容性目的,我保持了它的方式你有它.
在查找数组中,索引2(值为29)仅适用于闰年.
整个函数都很糟糕:它过于复杂,返回错误的结果,没有指定返回类型,并且不仅在参数名称上有类型前缀,而且它是错误的前缀.
这样做是这样的:
Function GetLastMonthDate(ByVal dtDateTime As DateTime) As DateTime
Return dtDateTime.AddDays(DateTime.DaysInMonth(dtDateTime.Year, dtDateTime.Month)-dtDateTime.Day)
End Function
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除此之外,看起来原始返回一个字符串,这实际上是不合适的.此函数应返回DateTime,并让调用它的函数处理相应的字符串转换.