How to identify sender and receiver id in inbound xml

Hi All,

Can any one of you please help me with the below requirement.

We will be receiving an inbound xml File through TN ,but the xml does not contain information of sender and receiver Id.

So for the inbound xml when received it selects the document type but there will not be sender and receiver to extract from that that case how can sender and receiver id be passed to that document type in TN

Hi Amruth,

How is the inbound xml being submitted to TN? Are you doing through some HTTP, FTP, AS2 or any other way? If AS2, you can utilize the AS2 Id as the unique id right?

Hi @Amruth_Ruttala,
In the DocType you can try giving the Xpath to define something Unique.
With AS2 ID you can definetly identify who is sending , but when the document gets published to Trading Network via EDIINT services we need to have something unique.
e.g In DocumentTypes
Identify Tab give the Xpath such as
/XMLDOCHEADER[0]/XMLDOC[0]/ELEMENTNAME[0]/FIELDNAME = (You can hard code here), this will help you to recognise the documentType if coming via AS2 and the first time when you receive the document it will be always EDIINT.
Also in Extract tab you can Add SenderID and give the XPATH , ReceiverID give the XPATH
This XPATH can be something unique that you have found.
Put that sender ID and Receiver ID information in the Trading partners so that this matches.
Hopefully this is what you were asking.
Please get back to me if my reply is not relevant.