Plan for today:

1

• Plan for today
• Quickly review the Data data type.
• Lean about some very useful date functions
• DatePart
• DateDiff
• DateAdd

2
There is a date/time type in VB
Date A Date variable can hold any date/time
from Jan 1, 100 to Dec 31, 9999
Dim dteEamonnsBday As Date dteEamonnsBday
4/4/1968
This is the American format
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The Date variable is really an object. Because
dates have multiple parts we may want to
access. Consider one of the greatest events in
human history, my birth. We may want to find out
its Year, day of week, day of month, day of
year, month, was it a leap year?, the minutes,
seconds, hours etc
So this generically is a date
Dim dteEamonnsBday As Date
And little modifiers like this
dteEamonnsBday.Year let us access parts of the
date
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Dim dteEamonnsBday As Date dteEamonnsBday
4/4/1968 DateDemo.Text dteEamonnsBday
Dim dteEamonnsBday As Date dteEamonnsBday
4/4/1968 DateDemo.Text dteEamonnsBday.ToLongDa
teString
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Dim dteEamonnsBday As Date dteEamonnsBday
4/4/1968 DateDemo.Text dteEamonnsBday.Year
DateDemo.Text dteEamonnsBday.DayOfWeek
DateDemo.Text dteEamonnsBday.Month
Suppose we want the text April, or the day
Thursday? (why does VB not do this
automatically?)
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One way to get the text for a day of week
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DateDemo.Text dteEamonnsBday.ToLongDateString
Alternative way to get the text for a day of week
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Dim dteToday As Date dteToday Now DateDemo.Text
dteToday
Now is a function, which takes the no
parameters, and returns the current date and time
DateDemo.Text dteToday.ToLongDateString
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We can access any part of a date with the
following syntax. In each case an integer is
returned
DateDemo.Text dteEamonnsBday.Month DateDemo.Text
dteEamonnsBday.Year DateDemo.Text
dteEamonnsBday.WeekOfYear DateDemo.Text
dteEamonnsBday.Quarter DateDemo.Text
dteEamonnsBday.DayOfWeek DateDemo.Text
dteEamonnsBday.Day DateDemo.Text
dteEamonnsBday.DayOfYear DateDemo.Text
dteEamonnsBday.Hour DateDemo.Text
dteEamonnsBday.Minute DateDemo.Text
dteEamonnsBday.Second
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Let us consider some useful functions for
manipulating dates
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You can also try the DatePart function instead of
any of the above functions. This function has 2
arguments, the first being a string corresponding
to what part of the date you want returned and
the other being the date expression. The DatePart
function can also return the quarter, the day of
the year, and the week of the year etc
DateDemo.Text "It is " DatePart("n", Now)
" past the hour."
Other acceptable strings to use for the first
argument are

• "yyyy" - identical to using Year function
• "m" - identical to using Month function
• "d" - identical to using Day function
• "w" - identical to using Weekday function
• ww - identical to using WeekOfYear function
• "h" - identical to using Hour function
• "n" - identical to using Minute function
• "s" - identical to using Second function

Why do this?
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Example of DatePart, timing a section of
code. How how does it take VB to count to a
billion?
Dim dteBegin, dteEnd As Date Dim
lngTimer, lngX As Long dteBegin Now
For lngX 1 To 1000000000 Next
dteEnd Now lngTimer
DatePart("s", dteEnd) - DatePart("s", dteBegin)
DateDemo.Text "It took "
lngTimer.ToString " seconds"
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The DateDiff function can tell you the difference
between two dates. Not just the number of days,
but any date or time interval. There are three
required arguments, the string corresponding to
the interval (these are the same strings listed
above for use with the DatePart function), and
the two dates to compare.
Dim dteEamonnsBday As Date Dim lngDays As
Long dteEamonnsBday "4/4/1968" lngDays
DateDiff("d", dteEamonnsBday, Now) DateDemo.Text
"My Age in days is " Str(lngDays)
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Dim dteNextStarWarsMovie As Date Dim lngDays As
Long dteNextStarWarsMovie "5/19/2005" lngDays
DateDiff("d", dteNextStarWarsMovie,
Now) DateDemo.Text Str(lngDays) " days until
the next SW movie."
We need to be careful with the order of the
paramenters
lngDays DateDiff("d", dteNextStarWarsMovie, Now)
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Alternative version of timing program How how
does it take VB to count to a billion?
Why different?
Dim dteBegin, dteEnd As Date Dim
lngTimer, lngX As Long dteBegin Now
For lngX 1 To 1000000000 Next
dteEnd Now lngTimer
DatePart("s", dteEnd) - DatePart("s", dteBegin)
lngTimer (DateDiff("s", dteBegin,
dteEnd)) DateDemo.Text "It took "
lngTimer.ToString " seconds"
Only change is here
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When comparing December 31 to January 1 of the
following year, DateDiff returns 1 for Year,
Month and quarter. In most cases this makes
sense for business and legal uses. However, if
you were speaking on New years day, you would not
say that a baby born the day before was 1 year
old. You need to be careful about the semantic,
cultural and legal meaning of date calculations.
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The DateAdd function can add or subtract date or
time intervals to a particular date. The first
argument is the string which represents the
interval (same strings as DatePart and DateDiff),
the second is the number of intervals to add or
subtract (positive numbers for future dates,
negative numbers for past dates), and the third
is the date to perform the calculation on.
When could a baby, born today, first exercise
their franchise?
Dim dteDOB, dteCanVote As Date dteDOB
Now dteCanVote DateAdd("yyyy", 18,
dteDOB) DateDemo.Text "You can vote on "
dteCanVote.ToLongDateString
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The DateAdd function is very intelligent. It
knows about leap years and it knows that all
months don't have the same number of days. For
example if you're trying to find the date one
month after Jan. 31, the function will return
Feb. 28 on non-leap years and Feb. 29 on leap
years.

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How do we detect leap years?
Dim lngYear As Long Dim sngRem As
Single lngYear DatePart("yyyy", Now)
sngRem (lngYear / 4) - Int(lngYear /
4) If sngRem gt 0 Then
DateDemo.Text "This is NOT a leap year"
Else DateDemo.Text "This is leap
year" End If
20
The previous code is actually incorrect!!!!
The year is defined as the length of time it
takes to pass from one vernal equinox to another.
If the calendar gains or loses days, the date for
the equinox shifts. Because the physical year
isn't exactly 365.25 days in length (as the
calendar says it should be), the current calendar
supplies 3 too many leap years every 385 years.
To make up for that, years divisible by 100
aren't leap years unless they're a multiple of
400. This means that 1700, 1800, and 1900
weren't leap years, but 2000 will be. The code
above would be wrong in 2100.
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Dim dteEamonnsBday As Date dteEamonnsBday
4/4/1968 DateDemo.Text dteEamonnsBday If
dteEamonnsBday.IsLeapYear(dteEamonnsBday.Year)
Then DateDemo.Text "The great sage Eamonn
was born on a leap year" End If
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If I were studying for the last quiz

• Could you write some code, that given an
  arbitrary year, say
• shtYear 2005
• Gives the date of new years day of that year?
• 2) Gives the date of thanksgiving of that year?
• 3) Calculates how many Tuesdays are in that
  April of that year?