Add support for DocumentFragment.getElementById()

Source/core/dom/ContainerNode.cpp

Index: Source/core/dom/ContainerNode.cpp
diff --git a/Source/core/dom/ContainerNode.cpp b/Source/core/dom/ContainerNode.cpp
index f355f618b05a5d2abd7d3807fca55973c52daf0d..07f4210845da661c3964387fb2ddcc35da445466 100644
--- a/Source/core/dom/ContainerNode.cpp
+++ b/Source/core/dom/ContainerNode.cpp
@@ -1162,6 +1162,26 @@ PassRefPtr<RadioNodeList> ContainerNode::radioNodeList(const AtomicString& name,
     return ensureRareData().ensureNodeLists().addCache<RadioNodeList>(*this, type, name);
 }
 
+Element* ContainerNode::getElementById(const AtomicString& id) const
+{
+    if (inDocument()) {

[arv (Not doing code reviews), 2014/04/25 13:36:19]

I don't think this is required. It should be sufficient to get the treeScope.

[Inactive, 2014/04/25 14:41:20]

Yes, this is not strictly required. I thought it was a nice addition though
because this optimization clearly does not make sense in the case we are not in
a Document (which is always the case for a DocumentFragment).

I think this avoids doing a bit of extra processing (hash look-up + checking if
element is a descendant) in the case the node *is not* in a Document and goes
straight to traversing the subtree.

Also, in the case the node *is* part of a Document, it avoids unnecessarily
traversing the subtree when treeScope().getElementById(id) returned null.

[Inactive, 2014/04/25 15:13:40]

I think I should be using isInTreeScope() instead of inDocument() since the
DocumentOrderedMap is associated to TreeScopes and not only Documents.

+        // Fast path if we are in a document: call getElementById() on document
+        // and check if the matching element is in our subtree.
+        Element* element = treeScope().getElementById(id);
+        if (!element)

[Inactive, 2014/04/25 14:41:20]

i.e. Early return here is now possible as we know we are in a Document and there
is no element with the given ID in the document (thus we avoid traversing).

[arv (Not doing code reviews), 2014/04/25 15:04:04]

I was thinking of the case where we are in a disconnected tree but the tree
scope knows about the IDs. Maybe that can never happen?

+            return 0;
+        if (element->isDescendantOf(this))
+            return element;
+    }
+
+    // Fall back to traversing our subtree. In case of duplicate ids, the first element found will be returned.
+    for (Element* element = ElementTraversal::firstWithin(*this); element; element = ElementTraversal::next(*element, this)) {
+        if (element->getIdAttribute() == id)
+            return element;
+    }
+    return 0;
+}
+
 #ifndef NDEBUG
 bool childAttachedAllowedWhenAttachingChildren(ContainerNode* node)
 {