DateDiff FunctionSee Also DateAdd Function | DatePart Function Requirements Version 2 Returns the number of intervals between two dates. DateDiff(interval, date1, date2 [,firstdayofweek[, firstweekofyear]]) The DateDiff function syntax has these parts:Arguments interval Required. String expression that is the interval you want to use to calculate the differences between date1 and date2. See Settings section for values. date1, date2 Required. Date expressions. Two dates you want to use in the calculation. firstdayofweek Optional. Constant that specifies the day of the week. If not specified, Sunday is assumed. See Settings section for values. firstweekofyear Optional. Constant that specifies the first week of the year. If not specified, the first week is assumed to be the week in which January 1 occurs. See Settings section for values. Settings The interval argument can have the following values:Setting Description yyyy Year q Quarter m Month y Day of year d Day w Weekday ww Week of year h Hour n Minute s Second The firstdayofweek argument can have the following values:Constant Value Description vbUseSystemDayOfWeek 0 Use National Language Support (NLS) API setting. vbSunday 1 Sunday (default) vbMonday 2 Monday vbTuesday 3 Tuesday vbWednesday 4 Wednesday vbThursday 5 Thursday vbFriday 6 Friday vbSaturday 7 Saturday The firstweekofyear argument can have the following values:Constant Value Description vbUseSystem 0 Use National Language Support (NLS) API setting. vbFirstJan1 1 Start with the week in which January 1 occurs (default). vbFirstFourDays 2 Start with the week that has at least four days in the new year. vbFirstFullWeek 3 Start with the first full week of the new year. Res You can use the DateDiff function to determine how many specified time intervals exist between two dates. For example, you might use DateDiff to calculate the number of days between two dates, or the number of weeks between today and the end of the year.To calculate the number of days between date1 and date2, you can use either Day of year ("y") or Day ("d"). When interval is Weekday ("w"), DateDiff returns the number of weeks between the two dates. If date1 falls on a Monday, DateDiff counts the number of Mondays until date2. It counts date2 but not date1. If interval is Week ("ww"), however, the DateDiff function returns the number of calendar weeks between the two dates. It counts the number of Sundays between date1 and date2. DateDiff counts date2 if it falls on a Sunday; but it doesn't count date1, even if it does fall on a Sunday.If date1 refers to a later point in time than date2, the DateDiff function returns a negative number.The firstdayofweek argument affects calculations that use the "w" and "ww" interval symbols.If date1 or date2 is a date literal, the specified year becomes a permanent part of that date. However, if date1 or date2 is enclosed in quotation s (" ") and you omit the year, the current year is inserted in your code each time the date1 or date2 expression is evaluated. This makes it possible to write code that can be used in different years.When comparing December 31 to January 1 of the immediately succeeding year, DateDiff for Year ("yyyy") returns 1 even though only a day has elapsed.The following example uses the DateDiff function to display the number of days between a given date and today: Function DiffADate(theDate) DiffADate = "Days from today: " & DateDiff("d", Now, theDate) End Function
DateAdd Function | DatePart Function
Requirements
Version 2
Returns the number of intervals between two dates. DateDiff(interval, date1, date2 [,firstdayofweek[, firstweekofyear]])
The DateDiff function syntax has these parts:Arguments
interval
Required. String expression that is the interval you want to use to calculate the differences between date1 and date2. See Settings section for values.
date1, date2
Required. Date expressions. Two dates you want to use in the calculation.
firstdayofweek
Optional. Constant that specifies the day of the week. If not specified, Sunday is assumed. See Settings section for values.
firstweekofyear
Optional. Constant that specifies the first week of the year. If not specified, the first week is assumed to be the week in which January 1 occurs. See Settings section for values.
Settings
The interval argument can have the following values:Setting Description
yyyy Year
q Quarter
m Month
y Day of year
d Day
w Weekday
ww Week of year
h Hour
n Minute
s Second The firstdayofweek argument can have the following values:Constant Value Description
vbUseSystemDayOfWeek 0 Use National Language Support (NLS) API setting.
vbSunday 1 Sunday (default)
vbMonday 2 Monday
vbTuesday 3 Tuesday
vbWednesday 4 Wednesday
vbThursday 5 Thursday
vbFriday 6 Friday
vbSaturday 7 Saturday The firstweekofyear argument can have the following values:Constant Value Description
vbUseSystem 0 Use National Language Support (NLS) API setting.
vbFirstJan1 1 Start with the week in which January 1 occurs (default).
vbFirstFourDays 2 Start with the week that has at least four days in the new year.
vbFirstFullWeek 3 Start with the first full week of the new year. Res
You can use the DateDiff function to determine how many specified time intervals exist between two dates. For example, you might use DateDiff to calculate the number of days between two dates, or the number of weeks between today and the end of the year.To calculate the number of days between date1 and date2, you can use either Day of year ("y") or Day ("d"). When interval is Weekday ("w"), DateDiff returns the number of weeks between the two dates. If date1 falls on a Monday, DateDiff counts the number of Mondays until date2. It counts date2 but not date1. If interval is Week ("ww"), however, the DateDiff function returns the number of calendar weeks between the two dates. It counts the number of Sundays between date1 and date2. DateDiff counts date2 if it falls on a Sunday; but it doesn't count date1, even if it does fall on a Sunday.If date1 refers to a later point in time than date2, the DateDiff function returns a negative number.The firstdayofweek argument affects calculations that use the "w" and "ww" interval symbols.If date1 or date2 is a date literal, the specified year becomes a permanent part of that date. However, if date1 or date2 is enclosed in quotation s (" ") and you omit the year, the current year is inserted in your code each time the date1 or date2 expression is evaluated. This makes it possible to write code that can be used in different years.When comparing December 31 to January 1 of the immediately succeeding year, DateDiff for Year ("yyyy") returns 1 even though only a day has elapsed.The following example uses the DateDiff function to display the number of days between a given date and today: Function DiffADate(theDate)
DiffADate = "Days from today: " & DateDiff("d", Now, theDate)
End Function
你说的对，但请问为什么请时间变量（而且时数据库TMSRS（XX））直接比较不行，非得加上一个CDATE（）函数呢？kokblack(人人为我，我为人人)
请说得具体点吗？yeefly(Web开发版)
你得英文版能不能针对我说得情况解释一下，能不能直接得到某月得最大天数？
tmsrs("start_date) < tmsrs("end_date") 比较得
难道，数据库ACCESS时间类型得段在ASP里是字符串？但一般将其用在一些函数参数都可以，难道那些函数都CDATE得强制转换。