$ZDATETIME($HOROLOG,3) - 3 hours

Question

Question

Michael Wood · Jan 30

#ObjectScript #InterSystems IRIS for Health

Anyone know how to subtract three hours from $ZDATETIME($HOROLOG,3)?

If $ZDATETIME($HOROLOG,3) returns 2025-01-30 10:17:04, I would like $ZDATETIME($HOROLOG,3) - 3 hours to return 2025-01-30 07:17:04

Product version: IRIS 2023.1

Discussion (7)3

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Matjaz Murko · Jan 30

Hi.

Try:
$SYSTEM.SQL.Functions.DATEADD("HH",-3,$H)

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Robert Cemper · Jan 31

this fails with <ILLEGAL VALUE> if time part is negative (before 3AM)

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Robert Cemper · Feb 25

slightly shorter using Julian Date

USER>set hours=-3 write $zdt($zdth(hours*3600+$zdt($h,-2),-2))

Sorry, I just couldn't resist

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score 0 · Answer 1 · 2025-02-26T04:29:29-05:00

This is the correct answer. $SYSTEM.SQL.Functions.DATEADD() will allow you add/subtract based on any of the date parts.

score 0 · Answer 2 · 2025-01-31T03:42:25-05:00

Ian Pears · Jan 31

you could try

w $ZDATETIME(+$H_","_($p($h,",",2)-10800),3)

there will be many ways.

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score 0 · Answer 3 · 2025-01-31T10:27:10-05:00

Ian Pears · Jan 31

so it would - thanks Robert

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score 0 · Answer 4 · 2025-02-25T15:19:26-05:00

You would need to manually roll back to the previous day to handle before 3am if you are subtracting 3 hours in seconds:

set newH = $H
set timeInSeconds = $PIECE(newH, ",", 2)
set timeInSeconds = timeInSeconds - (3 * 3600) ; Subtract 3 hours in seconds

if (timeInSeconds < 0) {
set newH = $PIECE(newH, ",", 1) - 1_","_(86400 + timeInSeconds) ; Roll back to previous day
} else {
set $PIECE(newH, ",", 2) = timeInSeconds
}

write $ZDATETIME(newH,3)