shijith.k
I have a DateTime with me, #DateTime<2019-11-10 13:27:00.0Z> and want to replace the time part of the DateTime struct with another time, say, 11:50:07.00+02:30. The solution I found was to convert the DateTime to string and then replace the time part using String.replace/3.
This is how my code looks.
dt = #DateTime<2019-11-11 13:27:00.0Z>
dt
|> DateTime.to_string()
|> String.replace(~r/[[:blank:]][[:alnum:]][[:alnum:]]:[[:alnum:]][[:alnum:]]:[[:alnum:]][[:alnum:]].[[:alnum:]]Z/, "T11:50:07.00+02:30")
|> DateTime.from_iso8601()
I feel there is a better way to do this, or at least there is a better regex to do it.
Does anyone have any better logic?
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mudasobwa
I would go with the direct struct update.
The above would discard the offset, though, and there is no clean way to get the offset in the
Timestruct because generally speaking the offset [arguably] has a meaning for the dates only. The below would be probable better:shijith.k
I can’t discard the offset, because after replacing the time, I still need to convert this to UTC.
So, for me, this is the suitable option. And its a much better and cleaner code than mine