Hi,

Try below command, this will give you the output in YYYYMMDDHHMM  format.  

w $extract($tr($zstrip("2022-03-29T15:10:00+0100","*p"),"TZ", ""),1,12)

Thanks,

Anusri

Hi,

Please try below, you should be able to read all json content. This works !

set oref = ##class(%Stream.GlobalCharacter).%New() do oref.Write(newMsg) // where newMsg contains JSON stream$$$LOGINFO("JSON = " _oref.Read())

You can also read only certain number of characters from json, for more understanding please go through below intersystems documentation Reading and Writing Stream Data

Thanks,

Anusri

Hello,

Please try below, this should help you !

write $zstrip($zdatetime($horolog, 3), "*pw") 

Output :  20220127113314

write $zstrip($zdatetime($horolog, 3), "*p") 

Output :  20220127113314

You can use either of the above two , both works !

Thanks,

Anusri

Extend your Message class with either Extends (%RegisteredObject, %JSON.Adaptor, %XML.Adaptor) or Extends (%Persistent, %XML.Adaptor, %JSON.Adaptor) .If your schemas line up then you can use below command to convert your message to json.

set tSC = tMessageObject.%JSONExportToString(.jsonEvent)

Informative ! Thanks for sharing !