Calendar dynamically should Take current year +1

Mon, 23 Nov 2015 07:33:23 GMT

Hi All,

i want to generate calendar to show current Year +1 ,how to do this in code.

for some Audit Reason they will upload future year data in advance ,it should be dynamic

LET Start = num(makedate(2014,1,1));

LET End =num(yearend(today()));

LET NumOfDays = End - Start + 1;

Date_src:

LOAD

$(Start) + Rowno() -1 as T_DateId

AUTOGENERATE $(NumOfDays);

Calendar:

LOAD

T_DateId as DateId,

T_DateId as Day,

year(T_DateId)*100+month(T_DateId) as MonthId,

year(T_DateId)*100+ceil(month(T_DateId)/3) as QuarterId,

month(T_DateId) as Month,

// dual(month(T_DateId) & '-' & year(T_DateId), year(T_DateId) & num(month(T_DateId), '00')) as MonthYear,

dual(month(T_DateId) & '-' & Date(T_DateId,'YY'), year(T_DateId)*100+month(T_DateId)) as MonthYear,

// dual(Date(T_DateId,'YY')&'-'&month(T_DateId), year(T_DateId)*100+month(T_DateId)) as MonthYear_YYMMM,

year(T_DateId) as Year,

'Q' & ceil(month(T_DateId)/3) as Quarter,

// dual('Q' & ceil(month(T_DateId)/3) & '-' & year(T_DateId), year(T_DateId) & ceil(month(T_DateId)/3)) as QuarterYear,

dual(Date(T_DateId,'YY ')&'Q' & ceil(month(T_DateId)/3) , year(T_DateId) & ceil(month(T_DateId)/3)) as QuarterYear,

Year(Today())-Year(T_DateId) as Year_Age,

(Year(Today())-Year(T_DateId))*12+(Month(Today())-Month(T_DateId)) as Month_Age,

(Year(Today())-Year(T_DateId))*12+(ceil(month(Today())/3)-ceil(month(T_DateId)/3)) as Quarter_Age

RESIDENT Date_src

;

Drop Table Date_src;

//DROP Table Calendar;

Re: Calendar dynamically should Take current year +1

Mark_Little — Mon, 23 Nov 2015 16:27:08 GMT

Hi,

Have you looked at the AddYears() function?

AddYears(Date,1) would add one year to the date field.

Mark

Tue, 24 Nov 2015 10:09:15 GMT

Hi Mark,

AddYears(yearend(today()),1) resolved issue

Thanks for your Suggestion.