To add 4 years to todays year and subtract 1 month present month

[Saikeerthana]

I want update

valid_year = todays year + 4

valid_month=todays month-1

I wrote the following code, but it is not working

var gd = new GlideDateTime(gs.nowDateTime());
//gdt=gd.getYearLocalTime();
gr.u_valid_year = gd.addYearsLocalTime(4);
gr.u_valid_month = gd.addMonthsLocalTime(1);

Thanks in advance!

[Hitoshi Ozawa]

Following code will add 4 years and substract 1 month to the current date.

var gdt = new GlideDateTime();
gdt.addYearsLocalTime(4);
gdt.addMonthsLocalTime(-1);
gs.print(gdt);

Execution result:

*** Script: 2026-01-20 09:18:00

[Saikeerthana]

how can I separate year and month and store in two different variable?

[Saurabh Gupta]

Hi,

Please try below code-

var gdt = new GlideDateTime();
gdt.addYearsLocalTime(4);
gdt.addMonthsLocalTime(-1);

gs.print("Date: "+gdt.getDate());
gs.print("Time: "+gdt.getTime().getDisplayValue());

If my answer replied your question please mark appropriate response as correct so that the question will appear as resolved for other users who may have a similar question in the future.

Regards,
Saurabh

[Saurabh Gupta]

Hi,

If my answer replied your question please mark appropriate response as correct so that the question will appear as resolved for other users who may have a similar question in the future.

Regards,
Saurabh

[Hitoshi Ozawa]

Following code will set variables month and year to a date that's 4 years from now and 1 month back.

var gdt = new GlideDateTime();
gdt.addYearsLocalTime(4);
gdt.addMonthsLocalTime(-1);
var month = gdt.getMonthLocalTime();
var year = gdt.getYearLocalTime();

gs.info("year:" + year + " month:" + month);

Execution result:

*** Script: year:2026 month:1

If prefer to have date in UTC instead of local time zone, use the following code:

var gdt = new GlideDateTime();
gdt.addYearsLocalTime(4);
gdt.addMonthsLocalTime(-1);
var month = gdt.getMonthUTC();
var year = gdt.getYearUTC();

gs.info("year:" + year + " month:" + month);