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# DATE2DWY Function

Member Posts: 53
Hello,
I am working on with date function date2dwy.
here my code
```MESSAGE(FORMAT(DATE2DWY(EndDate,1)));
```

So here my enddate is sunday and my message will display 7(Sunday) but my enddate will change sometimes and I always want to display 6 which is saturday. How can I do this dynamically. Let's say my enddate will be monday(1) and i want to change that to saturday(6).

Type: Date

The input date.

What
Type: Integer

Specifies what the function returns. The valid options are 1, 2, and 3.

The value 1 corresponds to day of the week (1-7, Monday = 1).

The value 2 corresponds to week number (1-53).

The value 3 corresponds to year.

• Member Posts: 53
Sorry it was a little bit complicated but I solved the problem by using a loop. The problem was enddate was always changing and I wanted to capture only one day. Thanks for your help!
```Day := DATE2DWY("12WEndDate",1);
WHILE DATE2DWY("12WEndDate",1) <> 6 DO
"12WEndDate" -= 1 ;
```
• Member Posts: 26
If you want to accomplish that you can use also
```"12WEndDate" := CALCDATE('<CW-1D>',"12WEndDate")
```

• Member Posts: 26
edited 2019-04-27
Maybe I don't understand what you want to do,
but when you want to display 6 in all circumstances, why don't you use

MESSAGE('6')
• Member Posts: 53
Sorry it was a little bit complicated but I solved the problem by using a loop. The problem was enddate was always changing and I wanted to capture only one day. Thanks for your help!
```Day := DATE2DWY("12WEndDate",1);
WHILE DATE2DWY("12WEndDate",1) <> 6 DO
"12WEndDate" -= 1 ;
```
• Member Posts: 26
```"12WEndDate" := CALCDATE('<CW-1D>',"12WEndDate")