DateDiff Function
Returns the number of date intervals between two given date values.
Синтакса
DateDiff (Add, Date1, Date2 [, Week_start [, Year_start]])
Врати вредност:
A number.
Параметри:
Add - A string expression from the following table, specifying the date interval.
Date1, Date2 - The two date values to be compared.
Week_start - An optional parameter that specifies the starting day of a week.
Week_start value |
Explanation |
0 |
Use system default value |
1 |
Sunday (default) |
2 |
Monday |
3 |
Tuesday |
4 |
Wednesday |
5 |
Thursday |
6 |
Friday |
7 |
Saturday |
Year_start - An optional parameter that specifies the starting week of a year.
Year_start value |
Explanation |
0 |
Use system default value |
1 |
Week 1 is the week with January, 1st (default) |
2 |
Week 1 is the first week containing four or more days of that year |
3 |
Week 1 is the first week containing only days of the new year |
Пример
Sub example_datediff
MsgBox DateDiff("d", "1/1/2005", "12/31/2005")
End Sub