Extract Message From JSON

[JJWest12]

I have a channel where the inbound message is a JSON object, and in that JSON object there is a node named "message" that contains an HL7v2 message. I am trying to extract the MSH.7 field from the HL7v2 message that is embedded in the JSON but cannot figure out how do to this. Can someone assist?

[kirbykn2]

What have you tried to do? Can you post an example TEST message (no PHI)

…

On Tue, Dec 10, 2024 at 8:44 AM JJWest12 ***@***.***> wrote: I have a channel where the inbound message is a JSON object, and in that JSON object there is a node named "message" that contains an HL7v2 message. I am trying to extract the MSH.7 field from the HL7v2 message that is embedded in the JSON but cannot figure out how do to this. Can someone assist? — Reply to this email directly, view it on GitHub <#6373>, or unsubscribe <https://github.com/notifications/unsubscribe-auth/APRXWDY3JDVBNO2CTNG2O4L2E3V4XAVCNFSM6AAAAABTLGN4VWVHI2DSMVQWIX3LMV43ERDJONRXK43TNFXW4OZXGY2TAMRVGA> . You are receiving this because you are subscribed to this thread.Message ID: ***@***.***>

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[pacmano1]

Stop replacing stuff. And stop .toString(). Stop setting the name space.

[pacmano1]

@JJWest12 don't blindly copy stuff without understanding what it does. Not giving you are hard time - but your code reflects that you don't understand what the code is doing.

[JJWest12]

@pacmano1 I need the replace because of the way the HL7 message in formatted in the JSON. It was "\r" where new lines need to be

[pacmano1]

\r is the standard delimiter for the end of a segment, not \n. In fact \n can be used inside a segment, e.g. in an OBX for text report.

@kirbykn2 already asked you for a sample message.

[pacmano1]

@JJWest12, I should have seen this eariler.

msg = SerializerFactory.getSerializer('HL7V2').toXML(msg['native_message'].toString().replace(/\\r/g, '\n'));

instead

msg = new XML(SerializerFactory.getSerializer('HL7V2').toXML(msg['native_message'].toString())