Checking to see if time is in a range

[cr84net2]

I am working on a reservation system and have never worked much with time in other applications. I am using military time to input "reservations" but I need to display when an item is reserved and was wondering if anybody knows a way to check if a given time is in a range between two other times: For Example:

If giventime is between rs("restimein") and rs("restimeout")
Then do something

Is there an easy way to do this, I haven't been able to find anything that seems to apply. I am going to need to check if about 32 different times fall between the reservation in and out range. (0700, 0730...2230, 2300)

Any help would be greatly appreciated.

 

[cr84net2]

I believe I have a function that will work if I can figure out how to convert a string to a time value. Any ideas?

 

[trollacious]

The DateDiff function is what you're looking for. From Microsoft:

Description
Returns the number of intervals between two dates.
Syntax
DateDiff(interval, date1, date2 [,firstdayofweek[, firstweekofyear]])
The DateDiff function syntax has these parts:

Part Description
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
vbUseSystem 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 weekof the new 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 a date literal, the specified year becomes a permanent part of that date. However, if date1 or date2 is enclosed in 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.

Lee