| 0 Votes |
YYYYMMDD of current and previous day
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| * Results in a "string"/number |
(concatenations of ((it as string) of year of it;(it as two digits) of month of it; (it as two digits) of day_of_month of it)) of (it; it - 1 * day) of current date
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| You are correct, fixed above. | |
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| I think "(it as two digits) of day_of_month of it" would provide the leading zero. With the above, March 9 is output as "2022039" | |
