Можно ли узнать НЕ сегодняшнюю дату?
Можно ли узнать число, которое будет или которое было (например, в прошлый или позапрошлый понедельник)? Или надо просто брать число, например, из Now и вычитать или прибавлять нужное кол-во дней + учитывать возможную смену месяцев (морока)? Т.е. нет ли возможно избегать этой довольно громоздкой арифметики и сразу получать нужную даты на заданном "расстоянии" от СЕГОДНЯ?
Для этого есть какие-то функции в объектной модели самого VB.Посмотри в модуле Date(так,кажись,его зовут) или поищи в Object Browser слово Date,а там и класс/модуль,за это ответственный, найдёшь
Эта функция - DateDiff
Вот описание из хелпа Вижуал Бейсика:
Returns a Variant (Long) specifying the number of time intervals between two specified dates.
Syntax
DateDiff(interval, date1, date2[, firstdayofweek[, firstweekofyear]])
The DateDiff function syntax has thesenamed arguments:
Part Description
interval Required.String expression that is the interval of time you use to calculate the difference between date1 and date2.
date1, date2 Required; Variant (Date). Two dates you want to use in the calculation.
firstdayofweek Optional. Aconstant that specifies the first day of the week. If not specified, Sunday is assumed.
firstweekofyear Optional. A 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.
Settings
The intervalargument has these settings:
Setting Description
yyyy Year
q Quarter
m Month
y Day of year
d Day
w Weekday
ww Week
h Hour
n Minute
s Second
The firstdayofweek argument has these settings:
Constant Value Description
vbUseSystem 0 Use the 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
Constant Value Description
vbUseSystem 0 Use the NLS API setting.
vbFirstJan1 1 Start with week in which January 1 occurs (default).
vbFirstFourDays 2 Start with the first week that has at least four days in the new year.
vbFirstFullWeek 3 Start with first full week of the year.
Remarks
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 adate literal, the specified year becomes a permanent part of that date. However, if date1 or date2 is enclosed in double quotation marks (" "), 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.
Вот описание из хелпа Вижуал Бейсика:
Returns a Variant (Long) specifying the number of time intervals between two specified dates.
Syntax
DateDiff(interval, date1, date2[, firstdayofweek[, firstweekofyear]])
The DateDiff function syntax has thesenamed arguments:
Part Description
interval Required.String expression that is the interval of time you use to calculate the difference between date1 and date2.
date1, date2 Required; Variant (Date). Two dates you want to use in the calculation.
firstdayofweek Optional. Aconstant that specifies the first day of the week. If not specified, Sunday is assumed.
firstweekofyear Optional. A 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.
Settings
The intervalargument has these settings:
Setting Description
yyyy Year
q Quarter
m Month
y Day of year
d Day
w Weekday
ww Week
h Hour
n Minute
s Second
The firstdayofweek argument has these settings:
Constant Value Description
vbUseSystem 0 Use the 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
Constant Value Description
vbUseSystem 0 Use the NLS API setting.
vbFirstJan1 1 Start with week in which January 1 occurs (default).
vbFirstFourDays 2 Start with the first week that has at least four days in the new year.
vbFirstFullWeek 3 Start with first full week of the year.
Remarks
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 adate literal, the specified year becomes a permanent part of that date. However, if date1 or date2 is enclosed in double quotation marks (" "), 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.
Спасибо за подсказки. По DateDiff в Object Browser'e нашел DateAdd, которая и дает число на заданном интервале от текущего. :)