(Ruby) XML Path Performance Optimizations

Discusses some important things to know about using Chilkat paths in the Chilkat XML API.

Chilkat Ruby Downloads install from rubygems.org gem install chilkat or download... Ruby Library for Windows, MacOS, Linux, Alpine Linux

require 'chilkat'

xml = Chilkat::CkXml.new()

# Let's load XML containing the following:

# <?xml version="1.0" encoding="utf-8"?>
# <xyz>
#     <licenses>
#         <license>
#             <id>1234</id>
#         </license>
#         <license>
#             <id>1234</id>
#         </license>
# ...
# My sample XML contains 64,000 "license" nodes ..
# ...
#         <license>
#             <id>1234</id>
#         </license>
#         <license>
#             <id>1234</id>
#         </license>
#     </licenses>
# </xyz>
# 
success = xml.LoadXmlFile("qa_output/large.xml")
if (success != true)
    print xml.lastErrorText() + "\n";
    exit
end

# Iterating over the individual "license" nodes with this code snippet is
# extremely slow:
licCount = xml.NumChildrenHavingTag("licenses|license")
print "license count = " + licCount.to_s() + "\n";

i = 0
# If "10" is changed to licCount, then it becomes apparent that this loop gets slower with each iteration.
while i < 10
    xml.put_I(i)
    s = xml.getChildContent("licenses|license[i]|id")
    print i.to_s() + ": " + s + "\n";
    i = i + 1
end

# The reason it is extremely slow is that the "license[i]" part of the path passed to GetChildContent
# says: find the i'th child of "licenses" having the tag "license".  Chilkat cannot assume that all
# children of an XML node have the same tag.  Therefore it's not possible to directly access the i'th child.
# Internally, Chilkat must start at the 1st child and iterate until it reaches the i'th child having the
# tag "license".

# For example, imagine if the XML was like this:

# <?xml version="1.0" encoding="utf-8"?>
# <xyz>
#     <licenses>
#         <license>
#             <id>1234</id>
#         </license>
#         <somethingElse>
#             <a>abc</a>
#         </somethingElse>
#         <license>
#             <id>1234</id>
#         </license>
# ...

# In the above XML, the 1st "license" is the 1st child of "licenses", but the 2nd "license"
# is the 3rd child of "licenses".

# If you already know that all children have the same tag, there is a shortcut that allows
# for direct access to that child.  Just leave off the tag name, like this:

i = 0
# If "10" is changed to licCount, then we can see the time for each loop is the same, and it's fast.
while i < 10
    xml.put_I(i)
    s = xml.getChildContent("licenses|[i]|id")
    print i.to_s() + ": " + s + "\n";
    i = i + 1
end

# When we pass just the index "[i]", we're saying: Get the i'th child regardless of tag.
# This is extremely fast because internally we can just access the i'th child directly.

# Another performance improvement is to call NumChildrenAt rather than NumChildrenHavingTag.
# For example:
licCount = xml.NumChildrenAt("licenses")
print "licCount = " + licCount.to_s() + "\n";

# NumChildrenAt returns the total number of children at the tag path.  If we already know
# all children will have the same tag, we can just get the count