Here is a snippet of code which is in use here: HapiContext hapiContext = new DefaultHapiContext(); hapiContext.setModelClassFactory(new CanonicalModelClassFactory(hl7Version)); hapiContext.setValidationRuleBuilder(new NoValidationBuilder());
final PipeParser pipeParser = hapiContext.getPipeParser(); final Message hapiMessage = pipeParser.parse(message); hapi library version 2.2 hl7Version is a string describing the hl7Version to use (we usually set it to “2.4”, you could set it to “2.3”, or you could use a default model class factory and let the pipe parser find the Hl7 version from the MSH). message is a string containing the hl7 message. Hope this helps Ian Systems Integration Team Qld Dept of Health Australia From: Chandan Datta [mailto:chandan.da...@auckland.ac.nz] Sent: Thursday, 30 October 2014 10:11 AM To: hl7api-devel@lists.sourceforge.net Subject: [HAPI-devel] Parse HL7 v2.3 REF message with local customizations Hi all, I am trying parse a HL7 REF I12 message with local customization(NZ). When I tried using the GenericParser, I keep getting Validation exceptions. For example for the segment below, I keep get the output ca.uhn.hl7v2.validation.ValidationException: Validation failed: Primitive value '(08)569-7555' requires to be empty or a US phone number PRD|PP|See T Tan^""^""^^""|""^^^^""^New Zealand||(08)569-7555||14134^NZMC My question is: * Is there a way to avoid the validation by using the conformance class generator * Is it possible to create own validation classes using CustomModelClasses? In either case, is there any example code for that or tutorial example documentation? -- Regards, Chandan Doctoral student,Robotics Research Group University of Auckland, New Zealand ******************************************************************************** This email, including any attachments sent with it, is confidential and for the sole use of the intended recipient(s). This confidentiality is not waived or lost, if you receive it and you are not the intended recipient(s), or if it is transmitted/received in error. Any unauthorised use, alteration, disclosure, distribution or review of this email is strictly prohibited. The information contained in this email, including any attachment sent with it, may be subject to a statutory duty of confidentiality if it relates to health service matters. If you are not the intended recipient(s), or if you have received this email in error, you are asked to immediately notify the sender by telephone collect on Australia +61 1800 198 175 or by return email. You should also delete this email, and any copies, from your computer system network and destroy any hard copies produced. If not an intended recipient of this email, you must not copy, distribute or take any action(s) that relies on it; any form of disclosure, modification, distribution and/or publication of this email is also prohibited. Although Queensland Health takes all reasonable steps to ensure this email does not contain malicious software, Queensland Health does not accept responsibility for the consequences if any person's computer inadvertently suffers any disruption to services, loss of information, harm or is infected with a virus, other malicious computer programme or code that may occur as a consequence of receiving this email. Unless stated otherwise, this email represents only the views of the sender and not the views of the Queensland Government. **********************************************************************************
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