DatePart: Difference between revisions

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m Ghenne moved page Datepart to DatePart
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DatePart returns a number specifying a part of the given date. The required parameter, ''interval'', determines the part of date which is measured and returned. The optional parameter ''firstdayofweek'' is Sunday, if not specified The optional ''firstweekofyear'' is the week containing January 1, if not specified.
DatePart returns a number specifying a part of the given date. The required parameter, ''interval'', determines the part of date which is measured and returned. The optional parameter ''firstdayofweek'' is Sunday, if not specified The optional ''firstweekofyear'' is the week containing January 1, if not specified.
'''Interval Values'''
{| class="wikitable"
|-
! Value !! 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
|}


== Example ==
== Example ==

Revision as of 11:42, 2 December 2012

DatePart(interval, date[, firstdayofweek[, firstweekofyear]])

Description

DatePart returns a number specifying a part of the given date. The required parameter, interval, determines the part of date which is measured and returned. The optional parameter firstdayofweek is Sunday, if not specified The optional firstweekofyear is the week containing January 1, if not specified.

Interval Values

Value 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

Example

Rem DatePart Example
'DatePart returns a numberfrompartofadate
Dim QuarterPart, MonthPart, DayPart
QuarterPart = DatePart("q", Now)
MonthPart = DatePart("m", Now)
DayPart = DatePart("d", Now)
Print "Today is day " & DayPart
Print "of month " & MonthPart
Print "in quarter " & QuarterPart

Output

Today is day 18
of month 8
in quarter 3
(''sample output from August 18, 1998'')

Related Items

DateAdd, DateDiff