days of week for a date

[fmtdonnk]

Hello

 

What is the function to get the week number of a certain date

 

for example I have a date

 

$sDate ="2018/05/5"

I would like to know if it's a Monday or Wednesday?

 

thanks

Franck

 

 

 

 

[fmtdonnk]

Hello

 

What is the function to get the week number of a certain date

 

for example I have a date

 

$sDate ="2018/05/5"

I would like to know if it's a Monday or Wednesday?

 

thanks

Franck

 

 

 

 

[JLogan3o13]

@fmtdonnk did you happen to notice the very clear statement at the top of the Examples forum? We even bolded it for you:

Quote

Do not post general support questions here, instead use the AutoIt Help and Support forums.

Please pay attention to where you post in the future.

As to your question - did you try looking in the help file? There is an entire section on Date Management where you'll find your answer pretty quickly.

[jchd]

Also note that your date isn't in the correct format for processing: it should have a leading zero for 1..9. Months & days need to have two digits each.

[fmtdonnk]

you mean under script and example?

By the way, finaly this is not working

I try this

$sDate ="2018/5/22"

Local $sLongDayName = _DateDayOfWeek($sNewDate)

but I get an error

when 

 

Local $sLongDayName = _DateDayOfWeek(@WDAY) it's working but I want to change the @WDAY 

 

THanks

 

 

 

[jchd]

Again, your date format is wrong.

[fmtdonnk]

Seems not wrong to me

 

in the same  script I have this , and it's working fine

==================================

$sDate ="2018/5/22"

For $x = 1 To 2
Local $sNewDate = _DateAdd('d', $x, $sDate)

MsgBox($MB_SYSTEMMODAL, "", "Today + 5 days: " &  $sNewDate)

 

Next

=================================

 

I see that the date increase by one day

 

I just need to check if it's Saturday or Sunday! and I stuck for 1 one hour for a stupid syntax I am sure from my side

 

[fmtdonnk]

thanks for your time

 

[jchd]

You're lucky single-digit month or day work in this case, but this point is clearly specified differently in the help file. That means things could stop working this way in some future release without notice (wouldn't be considered a script-breaking change). I for one wouldn't rely on this. Also date without leading zeroes (non-ISO) are a serious risk for sorting and interoperability.

[fmtdonnk]

ok I change and put a correct date format

==========================

$sDate ="2017/12/30"

For $x = 1 To 3
Local $sNewDate = _DateAdd('d', $x, $sDate)
Local $sLongDayName = _DateDayOfWeek($sDate)

MsgBox($MB_SYSTEMMODAL, "", "the date is : " &  $sNewDate & $sLongDayName)

 

Next

 

=============================

 

DateAdd working very good

but the DateDayOfWeek no.

I don't get any error, just don't nothing showing me the Week Day

 

I get

The date is : 2017/12/31

The date is : 2018/01/01

The date is : 2018/01/02

 

 

Do you have an idea how I can check if a date in the loop  is a Saturday or Sunday??

 

Thanks

 

Franck

[jchd]

You're misusing the _DateDayOfWeek function: it doesn't do what you believe it does.

Local $sDate ="2018/05/30"
Local $sNewDate, $aYMD, $iDoW, $sLongDayName
For $x = 0 To 3
    $sNewDate = _DateAdd('D', $x, $sDate)
    $aYMD = StringSplit($sNewDate, "/", 2)
    $iDoW = _DateToDayOfWeek($aYMD[0], $aYMD[1], $aYMD[2])
    $sLongDayName = _DateDayOfWeek($iDoW, 2)
    MsgBox($MB_SYSTEMMODAL, "", "The date is : " &  $sNewDate & " " & $sLongDayName)
Next

 

[Earthshine]

I found this in online help

https://www.autoitscript.com/autoit3/docs/libfunctions/_DateToDayOfWeekISO.htm

which will return numbers for days of week 1= Monday to 7 = Sunday, so, if it's 6 or 7 from this function that could be one way.

The example directly above works perfectly with either of these calls _DateToDayOfWeek and _DateToDayOfWeekISO

Edited May 30, 2018 by Earthshine

[jchd]

Both functions work but since the OP seemed to want the day longname as well in his code, it's preferable to use _DateToDayOfWeek  as it's Sunday-based just like _DateDayOfWeek. Of course if the goal is to decide if workday or not, then test for 6 or 7 is good enough.

[fmtdonnk]

Hello

 

Thanks for the link

 

but still same question 

Local $iWeekday = _DateToDayOfWeekISO(@YEAR, @MON, @MDAY)
How I put a date as a variable instead of @year @mon @Mday

 

=========================

$sDate ="2018/05/22"

Local $iWeekday = _DateToDayOfWeekISO($sDate)

====================

 

[fmtdonnk]

My question is how I can change into the function the date and put there a Variable that will increase by one into a loop

 

thanks

 

[jchd]

Did you look at the code I supplied?

[fmtdonnk]

HI JCHD

 

thank YOUUUUUUUU , No I didn't see your reply before. Now I tried it and its work perfectyl

 

I need to understand it, but before I do , I first thank you again

 

Have a very good evening

 

Franck