DateDiff - method of the Pm object
Description:
Returns the number of time-spans between two dates.
Syntax:
Long DateDiff(String timeSpan, Date date1, Date date2)
Parameters:
| timeSpan | (String) Specifies type of time span used to calculate the difference between date1 and date2
"q" - quarter
"w0" - week, where the first day is Sunday
"w1" - week, where the first day is Monday
"w2" - week, where the first day is Tuesday
"w3" - week, where the first day is Wednesday
"w4" - week, where the first day is Thursday
"w5" - week, where the first day is Friday
"w6" - week, where the first day is Saturday |
|---|
| date1 | (Date) First date in the calculation |
|---|
| date2 | (Date) Second date in the calculation |
|---|
Note:
This method is functional also in
Macro expression $.expr and in the
onDraw event of the
PmgCanvas object.
This method is also functional in
Web panels.
If is
date1 >
date2, then the function returns a negative number.
When comparing December 31 to January 1 of the immediately succeeding year, function for year (
"Y") returns 1 even if only a second has elapsed.
Difference between H, M, S and Hr, Mr, Sr:
If between
date1 and
date2 there is no transition standard time <-> daylight-saving time then there is no difference in those intervals.
If there is the transition then those intervals differ. In such case for example
"H" returns the calendar difference of hours and
"Hr" returns the real difference between those dates.
See
Example.
Example:
JavaScriptVBScriptSelect and copy to clipboard
var nVal;
var tDate = Pm.CreateDate(2018, 3, 25, 0, 0, 0, 0);
nVal = Pm.DateDiff("d", tDate, Pm.Time);
// Returns the number of days between a date tDate and today
// The following example presumes that on the 2018.3.25 there is transition from standard time to daylight-saving time
var tDate2 = Pm.CreateDate(2018, 3, 25, 5, 0, 0, 0);
nVal = Pm.DateDiff("H", tDate, tDate2);
// Returns 5
nVal = Pm.DateDiff("Hr", tDate, tDate2);
// Returns 4
Dim nVal
Dim tDate
tDate = Pm.CreateDate(2018, 3, 25, 0, 0, 0, 0)
nVal = Pm.DateDiff("d", tDate, Pm.Time)
' Returns the number of days between a date tDate and today
' The following example presumes that on the 2018.3.25 there is transition from standard time to daylight-saving time
Dim tDate2
tDate2 = Pm.CreateDate(2018, 3, 25, 5, 0, 0, 0)
nVal = Pm.DateDiff("H", tDate, tDate2)
' Returns 5
nVal = Pm.DateDiff("Hr", tDate, tDate2)
' Returns 4