Business days between date function

I use the datediff for business days function in a number of mysql data flows and it works great. However, I'm getting some very strange results on the latest flow that I'm working on and I can't figure out what's going wrong.

I have a start date which I'm looking to compare to the current date. However, the code I'm using is generating results that are greater than the number of normal days between the dates!

For example, the code below generates a value of 186 for the start date of Nov 26 when a simple datediff function generates 109. But there's no consistently: for other dates, lesser values are generated:


Any idea what's going on?!


PS: I know it's trivial to do this in ETL, so no big deal, it's just really bugging me!


Here's the code:


CASE WHEN 

`date_start` = CURRENT_DATE THEN 0 

ELSE

DATEDIFF(

CASE WHEN DAYOFWEEK(CURRENT_DATE) = 7 THEN SUBDATE(CURRENT_DATE, INTERVAL 1 DAY)

WHEN DAYOFWEEK(CURRENT_DATE) = 1 THEN SUBDATE(CURRENT_DATE, INTERVAL 2 DAY) ELSE CURRENT_DATE END

,

CASE WHEN DAYOFWEEK(`date_start`) = 1 THEN ADDDATE(`date_start`,INTERVAL 1 DAY)

WHEN DAYOFWEEK(`date_start`) = 7 THEN ADDDATE(`date_start`,INTERVAL 2 DAY) ELSE `date_start` END

-

((CASE WHEN WEEK(CASE WHEN DAYOFWEEK(`date_start`) = 1 THEN ADDDATE(`date_start`,INTERVAL 1 DAY)

WHEN DAYOFWEEK(`date_start`) = 7 THEN ADDDATE(`date_start`,INTERVAL 2 DAY) ELSE `date_start` END) = WEEK(CASE WHEN DAYOFWEEK(CURRENT_DATE) = 7 THEN SUBDATE(CURRENT_DATE, INTERVAL 1 DAY)

WHEN DAYOFWEEK(CURRENT_DATE) = 1 THEN SUBDATE(CURRENT_DATE, INTERVAL 2 DAY) ELSE CURRENT_DATE END) THEN 0 ELSE WEEK(CASE WHEN DAYOFWEEK(CURRENT_DATE) = 7 THEN SUBDATE(CURRENT_DATE, INTERVAL 1 DAY)

WHEN DAYOFWEEK(CURRENT_DATE) = 1 THEN SUBDATE(CURRENT_DATE, INTERVAL 2 DAY) ELSE CURRENT_DATE END) END


-

CASE WHEN WEEK(CASE WHEN DAYOFWEEK(`date_start`) = 1 THEN ADDDATE(`date_start`,INTERVAL 1 DAY)

WHEN DAYOFWEEK(`date_start`) = 7 THEN ADDDATE(`date_start`,INTERVAL 2 DAY) ELSE `date_start` END) = WEEK(CASE WHEN DAYOFWEEK(CURRENT_DATE) = 7 THEN SUBDATE(CURRENT_DATE, INTERVAL 1 DAY)

WHEN DAYOFWEEK(CURRENT_DATE) = 1 THEN SUBDATE(CURRENT_DATE, INTERVAL 2 DAY) ELSE CURRENT_DATE END) THEN 0 ELSE WEEK(CASE WHEN DAYOFWEEK(`date_start`) = 1 THEN ADDDATE(`date_start`,INTERVAL 1 DAY)

WHEN DAYOFWEEK(`date_start`) = 7 THEN ADDDATE(`date_start`,INTERVAL 2 DAY) ELSE `date_start` END) END) * 2)

END bd

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Best Answers

  • MarkSnodgrass
    MarkSnodgrass Portland, Oregon 🟤
    Accepted Answer

    One issue you are running into is using the week function and your dates cross years. Since the week numbers start over at the beginning of each year, you are going to have issues with the math. You would need to incorporate if the year of both dates are the same or not and make the necessary adjustments.

    I had to do a similar function and I started out starting small by just comparing two dates and trying to calculate the number of weekends between those two dates and then building my formula out from there.

  • GrantSmith
    GrantSmith Indiana 🟤
    Accepted Answer

    Hi @LCash

    Here's a simplified version for calculating the number of business days since today without using the WEEK function:

    -- Calculate the number of actual days between the two dates
    DATEDIFF(CURRENT_DATE(), `date_start`)
    -- Subtract the days of weekends between the two dates
    - DATEDIFF(CURRENT_DATE() + INTERVAL (1 - DAYOFWEEK(CURRENT_DATE())) DAY, `date_start` + INTERVAL (1 - DAYOFWEEK(`date_start`)) DAY) / 7 * 2
    -- If we start on a Sunday remove the partial week start
    - (CASE WHEN DAYOFWEEK(`date_start`) = 1 THEN 1 ELSE 0 END)
    -- If we end on a Saturday remove the partial end of the week.
    - (CASE WHEN DAYOFWEEK(CURRENT_DATE()) = 7 THEN 1 ELSE 0 END)
    
    

Answers

  • jaeW_at_Onyx
    jaeW_at_Onyx Budapest / Portland, OR 🟤

    SQL is a great tool, but I recommend you consider building a date dimension and then just marking the days as isWeekend or not.

    Today they want week days, tomorrow they'll ask you to exclude holidays.

  • LCash
    LCash ⚪️

    Thanks @MarkSnodgrass for the diagnosis, @GrantSmith for the solution, and @jaeW_at_Onyx for the recommendation!

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