Regex Replace conversion

[lledl]

HI,

I have date fiels that is in YYYY-MM-DD-HH.MM.SS.MSSSSS (EXAMPLE : 2016-11-16-10.48.12.123456)

How do I convert this 2016-11-16-10.48.12.123456 to 2016-11-16 10:48:12 with formula function

YYYY-MM-DD-HH.MM.SS.MSSSSS to YYYY-MM-DD HH:MM:S S

[RodL · Accepted solution]

You can typically use the Time/Date specifiers for just about any format. Check Help for the different specifiers available.

Try this expression...

DateTimeParse([DateTime], "%Y-%m-%d-%H.%M.%S")

Depending on your need for exact precision (which you eliminate anyway through any process), it looks like it truncates the milli-seconds as opposed to rounding though.