Ruby

XML Path Performance Optimizations

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

require 'chilkat'

success = false

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