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PaulMS
Super Contributor
This is similar to https://stackoverflow.com/questions/36179764/get-path-to-all-xmls-nodes
private static String getXPath(node) { if (node.parent().name() == node.name()) { return node.name() } return getXPath(node.parent()) + "/" + node.name() }
then use
output << getXPath(node)+'='+node.text()
evoks
7 years agoOccasional Contributor
Thanks a lot for your answer, that's exactly what I needed !
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