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# Adding days to a date

Posted on 2003-12-08
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I have a date field and i want to add days to it in a query or report.  I have a database that keeps track of people and dates for training and they need to be trained every so often. i want a report to pop up when the person needs to be trained again say in a year or two years

Here is how it looks:

John Smith        12/2/03

Now i want to add 365 days to that so it comes up with 12/2/04
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Question by:jpb12345
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Assisted Solution

Atropa earned 55 total points
Depends on what you want.

You can use DateAdd() to add  or subtract days, months or years.

do this..

that would add one year.  IF you do 365 next year it will be off since there is a leap year day so use what you want.

You can use this key to determine which to use..

yyyy    Year
q        Quarter
m       Month
y        Day of year
d        Day
w       Weekday
ww    Week
h       Hour
n       Minute
s       Second

Good luck.  /Atropa
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Author Comment

ok that worked perfectly like i wanted now how would i make it so if that new date = the current date then a msgbox appears or report comes up saying that these people need retraining
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Accepted Solution

shineyshoes earned 55 total points
Jpb12345 - Datevalue also works.
IE:

If datevalue([YourFieldName]) + 365) = (datevalue(now)) then
Msgbox "This person requires Retraining"
end if

3 Functions you might want to look up when working with dates are

DateValue()
DateDiff()

Very Usefull when working with Dates.

Shineyshoes
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LVL 5

Expert Comment

You would do this..

if DateAdd("y",1,[YourField]) = Date() then
MsgBox "This is today you idiot"
else
MsgBox "There are still " & DateDiff("d",DateAdd("y",1,[YourField]) ,Date()) & " left"
end if

What this will do is compare the date add with todays date.  Read this about DateDiff()..

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 these named 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. A constant 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 interval argument 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 a date 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.

Note   For date1 and date2, if the Calendar property setting is Gregorian, the supplied date must be Gregorian. If the calendar is Hijri, the supplied date must be Hijri.

/Atropa
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LVL 5

Expert Comment

Sorry shiney shoes, I didn't se your comment posted up there
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Author Comment

thanks for all the help

it works in the query fine and i see what your saying about the message box but i must be doing something wrong.  I'm just learning VBA.  If i want it to read right from the table or query dont i have to put the table's or query's name.  This is my idea:

I have a field called TrainingDate which is in a table called TBLEMPLOYEETRAINING.  Now from there i need to get the msgbox to take those dates for every record and tell me when they expire.  it keeps on giving me an error message saying that the database cant find the field stated in the expression.  What should i do
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