This function determines a specified part of a specified date.
DatePart(interval, date,[ firstdayofweek], [ firstweekofyear])
Setting |
Description |
yyyy | Year |
q | Quarter |
m | Month |
y | Day of year |
d | Day |
w | Weekday |
ww | Week of year |
h | Hour |
m | Minute |
s | Second |
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 |
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 week of the new year. |
Returns a specified part of a specified date.
Use DatePart to evaluate a date and return a specific interval of time. For example, you can use DatePart to calculate the day of the week or the current hour.
The firstdayofweek parameter affects calculations that use the w and ww interval symbols.
If date is a date literal, the specified year becomes a permanent part of that date. However, if date is enclosed in quotation marks ("") and you omit the year, the current year is inserted in your code each time the date expression is evaluated. This makes it possible to write code that can be used in different years.