使用dom4j定位具有行号的节点

Question

我使用dom4j来解析一个xml文件，我得到了一个带有行号的文档，我想使用行号来定位所有有行号的节点并操作这些节点。

是否有一种方法可以使用dom4j或其他dom4j来实现这一点？

Answer 1

可以将DOM4J扩展为包含元素的行号，但这有点复杂，而不是100%精确(您可以得到开始元素的">“字符的行号)。

可能，更健壮的方法是为每个违规元素报告一个XPath表达式，然后使用这些表达式来修复它。

不过，为了好玩，下面是一个完整的示例，说明如何在DOM4J中包含行号信息：

public class LineNumber {

    public static void main(String[] args) throws Exception {

        SAXReader reader = new MySAXReader();
        reader.setDocumentFactory(new LocatorAwareDocumentFactory());

        Document doc = reader
                .read(new StringReader("<root foo='bar' > \n<alfa/>\n<beta/>\n<gamma/>\n</root>\n"));
        doc.accept(new VisitorSupport() {
            @Override
            public void visit(Element node) {
                System.out.printf("%d: %s\n",
                        ((LocationAwareElement) node).getLineNumber(),
                        node.getName());
            }
        });

    }

    static class MySAXReader extends SAXReader {

        @Override
        protected SAXContentHandler createContentHandler(XMLReader reader) {
            return new MySAXContentHandler(getDocumentFactory(),
                    getDispatchHandler());
        }

        @Override
        public void setDocumentFactory(DocumentFactory documentFactory) {
            super.setDocumentFactory(documentFactory);
        }

    }

    static class MySAXContentHandler extends SAXContentHandler {

        private Locator locator;

        // this is already in SAXContentHandler, but private
        private DocumentFactory documentFactory;

        public MySAXContentHandler(DocumentFactory documentFactory,
                ElementHandler elementHandler) {
            super(documentFactory, elementHandler);
            this.documentFactory = documentFactory;
        }

        @Override
        public void setDocumentLocator(Locator documentLocator) {
            super.setDocumentLocator(documentLocator);
            this.locator = documentLocator;
            if (documentFactory instanceof LocatorAwareDocumentFactory) {
                ((LocatorAwareDocumentFactory)documentFactory).setLocator(documentLocator);
            }

        }

        public Locator getLocator() {
            return locator;
        }
    }

    static class LocatorAwareDocumentFactory extends DocumentFactory {

        private Locator locator;

        public LocatorAwareDocumentFactory() {
            super();
        }

        public void setLocator(Locator locator) {
            this.locator = locator;
        }

        @Override
        public Element createElement(QName qname) {
            LocationAwareElement element = new LocationAwareElement(qname);
            if (locator != null)
                element.setLineNumber(locator.getLineNumber());
            return element;
        }

    }

    /**
     * An Element that is aware of it location (line number in) in the source document
     */
    static class LocationAwareElement extends DefaultElement {

        private int lineNumber = -1;

        public LocationAwareElement(QName qname) {
            super(qname);
        }

        public LocationAwareElement(QName qname, int attributeCount) {
            super(qname, attributeCount);

        }

        public LocationAwareElement(String name, Namespace namespace) {
            super(name, namespace);

        }

        public LocationAwareElement(String name) {
            super(name);

        }

        public int getLineNumber() {
            return lineNumber;
        }

        public void setLineNumber(int lineNumber) {
            this.lineNumber = lineNumber;
        }

    }

}

Answer 2

检查器可以在节点中设置属性，将其标记为修复。这将是读取文件行的另一种选择。之后，只需查找具有此特定属性的节点并重构它。

<?xml version="1.0" encoding="UTF-8"?>
<tests>
    <test>okay</test>
    <test>good</test>
    <test>error</test>
</tests>

现在解析xml文件并标记最后一个要修复的测试节点。

public void parse(){

    SAXReader reader = new SAXReader();
    DocumentFactory documentFactory = DocumentFactory.getInstance();
    Document document = reader.read("test.xml");
    Element root = document.getRootElement();

    for (Iterator<Element> iterator = root.elementIterator();iterator.hasNext();){
        Element element = iterator.next();

        //check for your error and set fix flag, if not already happend 

        if(element.getText().equals("error") && 
           element.attributeValue("todo") == null) {

           Attribute fix = documentFactory.createAttribute(element, "todo" ,"fix");
            element.add(fix);
        }
    }

    // update xml file

    XMLWriter xmlWriter = new XMLWriter(new FileWriter("test.xml"));
    xmlWriter.write(document);
    xmlWriter.close();
}

输出xml

enter <?xml version="1.0" encoding="UTF-8"?>
<tests>
    <test>okay</test>
    <test>good</test>
    <test todo="fix">error</test>
</tests>