I do not know of any, and the official list of implementations doesn’t include one either. An alternative – of course less performant than a native implementation – would be XQIB which is an XQuery implementation in JavaScript. XPath 2.0 is fully included as a subset in XQuery 1.0, so you will be able to … Read more
General case The markup in the document following the root element must be well-formed. This error indicates that your XML has markup following the root element. In order to be well-formed, XML must have exactly one root element, and there can be no further markup following the single root element. One root element example (GOOD) … Read more
That “XML” is worse than invalid – it’s not well-formed; see Well Formed vs Valid XML. An informal assessment of the predictability of the transgressions does not help. That textual data is not XML. No conformant XML tools or libraries can help you process it. Options, most desirable first: Have the provider fix the problem … Read more
NSXMLParser is a SAX parser, meaning it steps through the XML document, and informs you (via delegate methods) of various events (such as starting an xml node, finding attributes, etc). This type of XML processing is best for parsing huge documents, and when you only need to retrieve a tiny amount of data from a … Read more
We had a similar issue – our head developer found a solution that works for us. We added this dependency to a couple of our pom.xml files For those that care, the unit tests in Sonar that were failing were apparently failing because Cobatura by default pulls in an old version of xerces. The version … Read more
DOM and Sax interfaces for XML parsing are the classic ways to work with XML. Python had to provide those interfaces because they are well-known and standard. The ElementTree package was intended to provide a more Pythonic interface. It is all about making things easier for the programmer. Depending on your build, each of those … Read more
XPath is right choice for you: DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance(); DocumentBuilder builder = factory.newDocumentBuilder(); Document doc = builder.parse(“<Your xml doc uri>”); XPathFactory xPathfactory = XPathFactory.newInstance(); XPath xpath = xPathfactory.newXPath(); XPathExpression expr = xpath.compile(“//Type[@type_id=\”4218\”]”); NodeList nl = (NodeList) expr.evaluate(doc, XPathConstants.NODESET); And iterate through nl
I think you want to do this: TimeSpan span = sunset-sunrise; TimeSpan half = new TimeSpan(span.Ticks / 2); DateTime result = sunrise + half; It can be written in one line if you want.
How to fix this issue ? Read the data using the correct character encoding. The error message means that you are trying to read the data as UTF-8 (either deliberately or because that is the default encoding for an XML file that does not specify <?xml version=”1.0″ encoding=”somethingelse”?>) but it is actually in a different … Read more
I had this problem as well. I did Project/Clean and the error went away and it works fine now.