有什么函数可以得到一个月的天数？

[woodheart]

比如8月31天
9月30天

谢了

[woodheart]

DateDiff(interval, date1, date2[, firstdayofweek[, firstweekofyear]])

返回两个日期之间的时间间隔数目

[masterkey]

我认为最简单的办法就是：
用下一个第一天减1就可以得到本月最后一天，比如要算8月份的天数，就以9月1日减1得到8月份31日，那么8月就是31天。
Dim vDate As Date
Dim mm, yy, dd As Integer
yy = Year(Date)
mm = Month(Date)
If mm = 12 Then
yy = yy + 1
mm = 1
Else
mm = mm + 1
End If
vDate = DateSerial(yy, mm, 1) - 1
dd = Day(vDate)

[wwwunix]

自已写一个嘛：
if (month in (1,3,5,7,8,10,12))
return 31;
else if (month in(4,6,9,11))
return 30;
else if (year 是闰年)
return 29;
else
return 28;
闰年的算法到处都有，就不写出来了。

[15年老程创业中]

VB里没有这样的函数，自己写一个吧。例如：

Public Function DaysOfMonth(Year As Integer, Month As Integer)
Dim Day1, Day2 As String
Day1 = CStr(Year) + "-" + CStr(Month) + "-1"
Day2 = CStr(Year) + "-" + CStr(Month + 1) + "-1"
DaysOfMonth = DateDiff("d", Day1, Day2)
End Function

[15年老程创业中]

VB里没有这样的函数，自己写一个吧。例如：

Public Function DaysOfMonth(Year As Integer, Month As Integer)
Dim Day1, Day2 As String
Day1 = CStr(Year) + "-" + CStr(Month) + "-1"
Day2 = CStr(Year) + "-" + CStr(Month + 1) + "-1"
DaysOfMonth = DateDiff("d", Day1, Day2)
End Function

[huntout]

我認為wwwunix的方法最直接最快！
閏年算法
((year mod 4 = 0) and (year mod 100 <> 0)) or (year mod 400 = 0)

[bing71]

DateDiff("d", "2000/05/01", "2000/06/01")可得出2000年5月的天数，以此类推。

[mxp]

to huntout:
ok!您的写法确实省去了不少不必要的循环。非常好！！
我是从以前作的打印月每天报表的程序中摘录的，思路被以前的需求束缚了，所以对于返回天数就没想到能进行优化。

呵呵！用下面的语句判断也可以：
Public Function DaysOfMonth(aYear As Integer, aMonth As Integer) As Integer
On Error Resume Next
For i = 31 To 28 Step -1
y = Month(CStr(aYear) & "-" & CStr(aMonth) & "-" & i)
If y = aMonth Then
DaysOfMonth = i
Exit Function
End If
Next
End Function

[youngeryang]

一行代码就行：
DateDiff("d",DateSerial(datYear,datMonth,1), DateAdd("m", 1,DateSerial(datYear,datMonth,1)))
其中datYear,datMonth是要计算的年和月。使用DateAdd可以保证计算12月时正确。

[mxp]

呵呵！还有一种方法的
Public Function DaysOfMonth(aYear As Integer, aMonth As Integer) As Integer
For i = 1 To 31
ine = IsDate(CStr(aYear) & "-" & CStr(aMonth) & "-" & i)
If ine = False Then
DaysOfMonth = i - 1
Exit Function
End If
Next
If ine = True Then DaysOfMonth = i - 1
End Function

[huntout]

to mxp:
想法不錯，但是你的程序好像有點問題唷！︰）

Public Function DaysOfMonth(aYear As Integer, aMonth As Integer) As Integer
Dim i As Integer
For i = 31 To 28 Setp -1
If IsDate(CStr(aYear) & "-" & CStr(aMonth) & "-" & i) Then
DaysOfMonth = i
Exit Function
End If
Next
End Function

[shines77]

不错，不错，想法不错，huntout改得很对！

[shines77]

nanman的代码最好应为：

Public Function DaysOfMonth(Year As Long, Month As Long) As Long
Dim Day1 As String, Day2 As String
Day1 = CStr(Year) + "-" + CStr(Month) + "-1"
Day2 = CStr(Year) + "-" + CStr(Month + 1) + "-1"
DaysOfMonth = DateDiff("d", Day1, Day2)
End Function

当然用变体的也对，但用Long锁定类型便于运算，减少出错。