From: Ian V. <Ian...@he...> - 2013-11-01 01:33:01
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As the use of HAPI extends deeper into our work practices, building test cases has encouraged us to use example message files. Saving them with the package, and bringing them in as resource streams is the manner we have adopted. Today I stumbled across the Hl7InputStreamMessageIterator, which looks to be made for just this scenario, so I have been having a go at using it. The code below shows my attempt: // Hl7InputStreamMessageIterator messageIterator = Hl7InputStreamMessageIterator.getForClasspathResource("/au/gov/qld/health/sit/cdr2/CdnPacsCbhInEscaped.dat"); Hl7InputStreamMessageIterator messageIterator = new Hl7InputStreamMessageIterator(getClass().getResourceAsStream("/au/gov/qld/health/sit/cdr2/CdnPacsCbhInEscaped.dat")); messageIterator.setIgnoreComments(true); // messageIterator.setMessageType(ca.uhn.hl7v2.model.v24.message.ORU_R01.class); As is I get a good result, and can step through the messages in my resource file and assert things for my test case. However, the commented out lines, the first being how I thought I might be able to bring in the resource, the second fixing the message to a particular version which I expect won't build. They return: Compiling 1 source file to D:\JavaDevelopment\RisToCdr2\build\test\classes D:\JavaDevelopment\RisToCdr2\test\au\gov\qld\health\sit\cdr2\CdnPacsCbhTxCdr2Test.java:52: error: cannot find symbol Hl7InputStreamMessageIterator messageIterator = Hl7InputStreamMessageIterator.getForClasspathResource("/au/gov/qld/health/sit/cdr2/CdnPacsCbhInEscaped.dat"); symbol: method getForClasspathResource(String) location: class Hl7InputStreamMessageIterator HAPI library in use is v2.1 Any thoughts on where I have gone wrong? Thanks Ian ******************************************************************************** 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. ********************************************************************************** |