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gonzl605
Regular Visitor

Send email -3 business day of the month

Hello - this might be a simple solution but it's driving me nuts. I want an automated email to be sent the 3rd business day before the end of the month. 

Example:

  • November 26th
  • January 27th
  • February 24th
1 ACCEPTED SOLUTION

Accepted Solutions
gonzl605
Regular Visitor

@DamoBird365 @HAZET @ScottShearer Thanks guys! I combined a bit from each of you to get to where I need to get to. Tested it by changing some info to make it true that today is the day I need and bam! Sent email. I've learned more about Power Automate in this thread than I have in the last month!

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12 REPLIES 12
ScottShearer
Super User
Super User

@gonzl605 

There are really a number of parts to your question.  First, you need to find the end of the current month.

Here is a blog post to find the end of the month

Once you find the end of the month, you can use addDays() and subtract 2 more days from it.

Then you need to know if the resulting date is a weekday.

Use a dayOfWeek() expression - that will return Sunday as 0 and Saturday as 6.

You can use a condition or write an expression that says if the day of week is 0 or 6, subtract two days from the date (use an addDays() expression).

 

If you need an example, let me know.

 

If I have answered your question, please mark your post as Solved.
If you like my response, please give it a Thumbs Up.

Scott

Hi @gonzl605 

 

I notice @ScottShearer has given you an idea, I was working on the solution and will go ahead and post my version as follows.

 

First, using the "Add to Time" action, I can get the start of this month and add 1 month to the date i.e. with today being 09/11/2021 (it the UK at least 😅), this would return 01/12/2021.  Then using the same action again, I am able to subtract a day and get the end of this month i.e. 30/11/2021.

 

DamoBird365_1-1636490601858.png

Next I have created an array of dates (using a select on a range 0 to 7, the last 7 days of this month and used an expression to indicate the day of the week, 0 and 6 being Sunday and Saturday.  I use sub to turn the item() into a negative, i.e. Sub(0,0) is 0, sub(0,1) is -1, then -2 etc.  This allows me to get the last 7 days of the month, including the last day of the month.  I probably on need the last 5 but this will allow you or someone else to adapt for 4th last day, 5th last day etc.


Select:

FROM range(0,7)

date adddays(body('Last_Day_of_This_Month'),sub(0,item()))

dayofweek dayOfWeek(adddays(body('Last_Day_of_This_Month'),sub(0,item())))

 

Filter Array (in advanace mode)

@and(not(equals(item()?['dayofweek'], 0)),not(equals(item()?['dayofweek'], 6)))

 

DamoBird365_2-1636490702116.png

The final compose is to return the 3rd element of the array (after it has been filtered).  Elements start from 0,1,2 and so the 3rd element is called by integer 2.

 

body('Filter_array')?[2]?['date']

 

Hopefully this helps you to achieve your goal, pretty complicated but a good challenge.

 

Please consider accepting my answer as a solution if it helps to solve your problem.

Cheers
Damien

Please take a look and subscribe to my YouTube Channel for more Power Platform ideas and concepts, or take a look at my website. Thanks

 

HAZET
Regular Visitor

To find the day three days before the month ends, add a condition and compare 

startOfDay(utcNow())

to

subtractFromTime(startOfMonth(addToTime(utcNow(),1,'Month')),3,'Day')

 

The first expression just cuts the hours, minutes and seconds off the timestamp, so you get the plain date.

The second adds a month to the timestamp which will move you into the following month (don't worry, if you add a month to the 1st of January, you will get the 28th of February, not the 2nd of March), then goes to the start of this following month and subtracts three days, which brings you back three days into the current month from the end of it.

 

If both render the same result, you know you are three days before the current month ends.

@DamoBird365 Thanks a ton for the solution! As a newbie it took me a while to follow but I got it to work! 🙂 

Would you mind finishing it off and letting me know how I loop it in with actually sending the email? I am familiar with the email composure piece but how do I get to to shoot off the email on that specific date? Is it as simple as adding "Send email with options"?

Hi @gonzl605 

 

I would setup a daily recurrence trigger, and then implement a condition.  If utcnow() equals 3rd business day of month, branch yes, else branch no.  You might need to use formatdatetime() to ensure both dates are in same format.  I.e. formatdatetime(date,'yyyy-MM-dd').

 

I guess the other option would be to set up a recurrence based on 22nd of each month and then use the delay action based on the new date.  Your flow would wait until that date and wouldn't therefore run daily.

 

Damien

HAZET
Regular Visitor

Hi,

to achieve a mail being sent off three workdays before the end of a month, you'd have to do several things. Parts are already described above.

  • Create a scheduled cloud flow that is triggered once every day
  • Make the action conditional, i. e. the sending of the mail. The condition would be that the current date matches the criterion "three workdays before the end of the month". On all but one day of each month your flow won't do anything.
  • Design the criterion properly. What do you want to take into account:
    • Weekdays, i. e. Saturday and Sunday don't count. This is pretty regular and can be handled using a formula
    • Holidays - difficult, the patterns aren't that easy to model. You'd probably need a comparison to some kind of calendar

Can you maybe be more specific regarding the latter issue? Holidays or not?

 

Best

Henrik

HAZET
Regular Visitor

I thought about the "business" day question a little longer. As long as we can ignore holidays, a formula calculating the third-last business day of the current month is:

 

 

subtractFromTime(startOfMonth(addToTime(utcNow(),1,'Month')),min(4,max(2,mod(add(4,dayOfWeek(utcNow())),7))),'Day')

 

 

HAZET
Regular Visitor

HAZET_0-1636540884487.png

 

The condition is 

startOfDay(utcNow()) eq subtractFromTime(startOfMonth(addToTime(utcNow(),1,'Month')),min(4,max(2,mod(add(3,dayOfWeek(utcNow())),7))),'Day')

 

Hi @HAZET 

 

I like that you have used expressions and a single action but I am not getting the 3rd last business day of the month.

 

subtractFromTime(startOfMonth(addToTime(utcNow(),1,'Month')),min(4,max(2,mod(add(4,dayOfWeek(utcNow())),7))),'Day')

 

Can you explain your thinking as it is definitely much cleaner than mine 😉

 

Damien

HAZET
Regular Visitor

Hi Damien,

I shall try 🙂

  • I first look at the pattern, how does the weekend affect the effective number of days you have to go back
    • If the last of the month is Sunday to Tuesday, you have to go 4 days back (e. g. 0 Tuesday, -1 Monday, -2 Sunday, -3 Saturday, -4 Friday)
    • If the last of the month is a Saturday, only 3 days: 0 Saturday, -1 Friday, -2 Thursday, -3 Wednesday
    • If it is Wednesday to Friday, two days back suffice
  • Then I looked for a function that would give me those numbers
    • dayofweek(utcNow()) gives you the day of the week as 0 for Sunday to 6 for Saturday
    • mod(add(4,dayofweek(utcNow())),7) shifts this row by 4, so you get 0 for Wednesday to 6 for Tuesday, in particular a 2 for Friday, a 3 for Saturday and a 4 for Sunday. Looks nice already, but the numbers for the other days are either too large or too small.
    • min(2,max(4,mod(add(4,dayofweek(utcNow())),7))) makes sure you get at least a 2 but no more than 4. The row should now be Su 4, Mo 4, Tu 4, We 2, Th 2, Fr 2, Sa 3 - which is exactly what we need.
  • now we calculate the first day of the following month startOfMonth(addToTime(utcNow(),1,'Month')) 
  •  then we go back from there the calculated number of days - and here is where I made a mistake i fear - I need to shift the whole row one day back, as I wanted to get three workdays including the last of the month, now I am getting three days from the beginning of the next month.

Should be easily fixed - I just don't have the time to test it right now. I assume shifting min, max and add up one each will do the trick.

gonzl605
Regular Visitor

@DamoBird365 @HAZET @ScottShearer Thanks guys! I combined a bit from each of you to get to where I need to get to. Tested it by changing some info to make it true that today is the day I need and bam! Sent email. I've learned more about Power Automate in this thread than I have in the last month!

Hi @HAZET 

 

I like the use of Mod - clever.  I was in the process of working it out, my notepad has 4 4 4 2 2 2 3 - just like you - definitely a different approach but more efficient!  I see that @gonzl605 is sorted.  Thanks for sharing.

 

Damien

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