Lua

XML Path Performance Optimizations

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

local chilkat = require("chilkat")

local xml = chilkat.Xml.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>
--
local success = xml:load_xml_file("qa_output/large.xml")
if not success then
    print(xml.last_error_text)
    return
end

-- Iterating over the individual "license" nodes with this code snippet is
-- extremely slow:
local licCount = xml:num_children_having_tag("licenses|license")
print("license count = " .. licCount)

local s = ""
local i = 0
-- If "10" is changed to licCount, then it becomes apparent that this loop gets slower with each iteration.
while i < 10 do
    xml.i = i
    s = xml:get_child_content("licenses|license[i]|id")
    print(i .. ": " .. s)
    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 do
    xml.i = i
    s = xml:get_child_content("licenses|[i]|id")
    print(i .. ": " .. s)
    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:num_children_at("licenses")
print("licCount = " .. licCount)

-- 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.