topic Re: Let year(today) problem showing todays result as today in App Development

Let year(today) problem showing todays result as today

richards — Mon, 07 Dec 2020 10:15:14 GMT

In my application i´m loading the year 2019, 2020, 2021 from the database.

In my application i would like to use my own named variabel CurrentYear, so that i could divide the values between this year and last years. etc....

However in the database there are more than three years so i´m using the following statement in the script.

Let vMinYear = Year(Today))+1;

Let vMaxYear = Year(Today))-1;

On the database i´m then determing the year as i = $(vMaxYear) To $(vMinYear) so it´s only loading three years.

However when making a +1 statement the 2021 is determined as a currentYear.

So the questions is how to dermine so that the current year is 2020 and to make it possible to choose other years

The currentYear should always be today.

Re: Let year(today) problem showing todays result as today

Lisa_P — Tue, 08 Dec 2020 03:50:31 GMT

It appears you have your Min and Max reversed.

Re: Let year(today) problem showing todays result as today

richards — Tue, 08 Dec 2020 06:42:14 GMT

Yes of course. However this doesn´t explain why the currentYear Value that is shown is 2021. I´m woundering if this could be changed so the data that is shown is today.

The max and min is only used to load number of years.

Re: Let year(today) problem showing todays result as today

tresB — Tue, 08 Dec 2020 07:00:39 GMT

@richards

I guess I didn't get it right. What do you get for year(today()) ?

Re: Let year(today) problem showing todays result as today

richards — Tue, 08 Dec 2020 07:07:48 GMT

on the variable list in the arc i have current year as defined as =Max(Year)

The varables vMaxYear is 2021 now after i rearanged the min and max. i.e. now vMinYear 2020.

Year today is 2020.