compareDocumentPosition method

DocumentPosition compareDocumentPosition(Node other)

Implementation

DocumentPosition compareDocumentPosition(Node other) {
  if (this == other) {
    return DocumentPosition.EQUIVALENT;
  }

  // We need to find a common ancestor container, and then compare the indices of the two immediate children.
  List<Node> chain1 = [];
  List<Node> chain2 = [];
  Node? current = this;
  while (current != null && current.parentNode != null) {
    chain1.insert(0, current);
    current = current.parentNode;
  }
  current = other;
  while (current != null && current.parentNode != null) {
    chain2.insert(0, current);
    current = current.parentNode;
  }

  // If the two elements don't have a common root, they're not in the same tree.
  if (chain1.first != chain2.first) {
    return DocumentPosition.DISCONNECTED;
  }

  // Walk the two chains backwards and look for the first difference.
  for (int i = 0; i < math.min(chain1.length, chain2.length); i++) {
    if (chain1[i] != chain2[i]) {
      if (chain2[i].nextSibling == null) {
        return DocumentPosition.FOLLOWING;
      }
      if (chain1[i].nextSibling == null) {
        return DocumentPosition.PRECEDING;
      }

      // Otherwise we need to see which node occurs first.  Crawl backwards from child2 looking for child1.
      Node? previousSibling = chain2[i].previousSibling;
      while (previousSibling != null) {
        if (chain1[i] == previousSibling) {
          return DocumentPosition.FOLLOWING;
        }
        previousSibling = previousSibling.previousSibling;
      }
      return DocumentPosition.PRECEDING;
    }
  }
  // There was no difference between the two parent chains, i.e., one was a subset of the other.  The shorter
  // chain is the ancestor.
  return chain1.length < chain2.length ? DocumentPosition.FOLLOWING : DocumentPosition.PRECEDING;
}