Replace time in a DateTime struct

Questions / Help
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1

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?

2

I would go with the direct struct update.

with {:ok, dt, 0} <- DateTime.from_iso8601("2019-11-10 13:27:00.0Z"),
     {:ok, t} <- Time.from_iso8601("11:50:07.00+02:30"),
  do: %DateTime{dt | hour: t.hour, minute: t.minute,
                     second: t.second, microsecond: t.microsecond}
#⇒ ~U[2019-11-10 11:50:07.00Z]

The above would discard the offset, though, and there is no clean way to get the offset in the Time struct because generally speaking the offset [arguably] has a meaning for the dates only. The below would be probable better:

date =
  DateTime.from_iso8601("2019-11-10 13:27:00.0Z")
  |> elem(1)
  |> DateTime.to_date()
  |> Date.to_iso8601()
DateTime.from_iso8601(date <> " 11:50:07.00+02:30")                      
#⇒ {:ok, ~U[2019-11-10 09:20:07.00Z], 9000}
2 Likes
3

The above would discard the offset

I can’t discard the offset, because after replacing the time, I still need to convert this to UTC.

date =
DateTime.from_iso8601(“2019-11-10 13:27:00.0Z”)
|> elem(1)
|> DateTime.to_date()
|> Date.to_iso8601()
DateTime.from_iso8601(date <> " 11:50:07.00+02:30")
#⇒ {:ok, ~U[2019-11-10 09:20:07.00Z], 9000}

So, for me, this is the suitable option. And its a much better and cleaner code than mine :smile: