Пример #1
0
 /**
  * Returns the value of the specified attribute or {@code null}.
  *
  * @param name attribute to be found
  * @return attribute value
  */
 public byte[] attribute(final QNm name) {
   final BasicNodeIter iter = attributes();
   while (true) {
     final ANode node = iter.next();
     if (node == null) return null;
     if (node.qname().eq(name)) return node.string();
   }
 }
Пример #2
0
 /**
  * Recursively finds the uri for the specified prefix.
  *
  * @param pref prefix
  * @return uri
  */
 public final byte[] uri(final byte[] pref) {
   final Atts at = namespaces();
   if (at != null) {
     final byte[] s = at.value(pref);
     if (s != null) return s;
     final ANode n = parent();
     if (n != null) return n.uri(pref);
   }
   return pref.length == 0 ? Token.EMPTY : null;
 }
Пример #3
0
  /**
   * Parses a string as XML and adds the resulting nodes to the specified parent.
   *
   * @param value string to parse
   * @param elem element
   */
  public static void add(final byte[] value, final FElem elem) {

    try {
      final Parser parser = new XMLParser(new IOContent(value), MainOptions.get(), true);
      for (final ANode node : new DBNode(parser).children()) elem.add(node.copy());
    } catch (final IOException ex) {
      // fallback: add string representation
      Util.debug(ex);
      elem.add(value);
    }
  }
Пример #4
0
 /**
  * Returns a copy of the namespace hierarchy.
  *
  * @param sc static context (can be {@code null})
  * @return namespaces
  */
 public final Atts nsScope(final StaticContext sc) {
   final Atts ns = new Atts();
   ANode node = this;
   do {
     final Atts nsp = node.namespaces();
     if (nsp != null) {
       for (int a = nsp.size() - 1; a >= 0; a--) {
         final byte[] key = nsp.name(a);
         if (!ns.contains(key)) ns.add(key, nsp.value(a));
       }
     }
     node = node.parent();
   } while (node != null && node.type == NodeType.ELM);
   if (sc != null) sc.ns.inScope(ns);
   return ns;
 }
Пример #5
0
 /**
  * Compares two nodes for their unique order.
  *
  * @param node1 first node
  * @param node2 node to be compared
  * @return {@code 0} if the nodes are identical, or {@code 1}/{@code -1} if the first node appears
  *     after/before the second
  */
 static int diff(final ANode node1, final ANode node2) {
   // cache parents of first node
   final ANodeList nl = new ANodeList();
   for (ANode n = node1; n != null; n = n.parent()) {
     if (n == node2) return 1;
     nl.add(n);
   }
   // find lowest common ancestor
   ANode c2 = node2;
   LOOP:
   for (ANode n = node2; (n = n.parent()) != null; ) {
     final int is = nl.size();
     for (int i = 1; i < is; i++) {
       if (n == node1) return -1;
       if (!nl.get(i).is(n)) continue;
       // check which node appears as first LCA child
       final ANode c1 = nl.get(i - 1);
       final BasicNodeIter ir = n.children();
       for (ANode c; (c = ir.next()) != null; ) {
         if (c.is(c1)) return -1;
         if (c.is(c2)) return 1;
       }
       break LOOP;
     }
     c2 = n;
   }
   // subtraction is used instead of comparison to support overflow of node id
   return node1.id - node2.id < 0 ? -1 : 1;
 }
Пример #6
0
 @Override
 public Item item(final QueryContext qc, final InputInfo ii) throws QueryException {
   final ANode node = toEmptyNode(arg(0, qc), qc);
   final QNm qname = node != null ? node.qname() : null;
   return qname != null ? Uri.uri(qname.uri(), false) : Uri.EMPTY;
 }
Пример #7
0
 /**
  * Adds children of a sub node.
  *
  * @param ch child nodes
  * @param nb node cache
  */
 static void addDesc(final BasicNodeIter ch, final ANodeList nb) {
   for (ANode n; (n = ch.next()) != null; ) {
     nb.add(n.finish());
     addDesc(n.children(), nb);
   }
 }
Пример #8
0
 /**
  * Returns the root of a node (the topmost ancestor without parent node).
  *
  * @return root node
  */
 public final ANode root() {
   final ANode p = parent();
   return p == null ? this : p.root();
 }