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Parsing Datetime with symbols

KellyT

Dear community:

I am having trouble to find a good solution to parse Datetime examples below by changing the "Format" (under "Column Info"):

2018-01-01 14:44:19.290000+00:00

2018-01-01 13:43:40.090000+00:00

Currently, I have to use Regex to find the pattern after the period and then replace it with blank. I am hoping that there is an easy option.

Thank you for your help!

Accepted answers

Mark_Bailey

I don't know if this result is quite what you want but it is simple.

I used this formula in another column and left the original alone.

Informat( Substitute( Word( 1, :Date Time String, "+" ), " ", "T" ), "yyy-mm-dd" )

All comments

Jeff_Perkinson

Again, my favorite function comes the rescue. Word(), using "+" as the delimiter can give you the bit before the plus. Then you can use Informat() to convert it to a date value. Informat(), without a second argument, will let JMP figure out the format on it's own if it can.

2020-08-13_12-38-41.531.png

Read more about these two topics:

If you learn only one Formula Editor function, Word() is the one

Using dates, times, datetimes and durations in JMP

Mark_Bailey

I don't know if this result is quite what you want but it is simple.

I used this formula in another column and left the original alone.

Informat( Substitute( Word( 1, :Date Time String, "+" ), " ", "T" ), "yyy-mm-dd" )

Craige_Hales

The T format is really useful for exchanging date time values, even if it is a bit ugly.

https://en.wikipedia.org/wiki/ISO_8601#Combined_date_and_time_representations