Lua

Parsing XML Children Having Identical Tags

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Demonstrates how to deal with XML where child elements may have identical tags. Lua

local chilkat = require("chilkat")

-- This example will get the phoneNumber's and groupMembershihpInfo's from the following XML:

-- <?xml version="1.0" encoding="UTF-8" ?>
-- <someEntries>
--     <entry>
--         <id>123</id>
--         <updated>2017-07-19T05:19:57.761Z</updated>
--         <app:edited xmlns:app="http://www.w3.org/2007/app">2017-07-19T05:19:57.761Z</app:edited>
--         <category scheme="http://schemas.google.com/g/2005#kind" term="http://schemas.google.com/contact/2008#contact"/>
--         <title>George Costanza</title>
--          <gd:name>
--             <gd:fullName>George Costanza</gd:fullName>
--             <gd:givenName>George</gd:givenName>
--             <gd:familyName>Costanza</gd:familyName>
--         </gd:name>
--         <gd:phoneNumber rel="http://schemas.google.com/g/2005#home">(555) 123-4567</gd:phoneNumber>
--         <gd:phoneNumber rel="http://schemas.google.com/g/2005#mobile">(555) 444-8877</gd:phoneNumber>
--         <gd:phoneNumber rel="http://schemas.google.com/g/2005#work">(555) 678-1111</gd:phoneNumber>
--         <gContact:groupMembershipInfo deleted="false" href="http://www.google.com/123"/>
--         <gContact:groupMembershipInfo deleted="false" href="http://www.google.com/456"/>
--     </entry>
-- </someEntries>
--

local xml = chilkat.Xml.new()

local success = xml:load_xml_file("qa_data/xml/georgeCostanza.xml")

local numPhoneNumbers = xml:num_children_having_tag("entry|*:phoneNumber")
local i = 0
while i < numPhoneNumbers do
    local xPhoneNumber = xml:get_nth_child_with_tag("entry|*:phoneNumber", i)
    print(xPhoneNumber.content)

    i = i + 1
end

print("----")

local numGroupMemberships = xml:num_children_having_tag("entry|*:groupMembershipInfo")
i = 0
while i < numGroupMemberships do
    local xMembership = xml:get_nth_child_with_tag("entry|*:groupMembershipInfo", i)
    print(xMembership:get_attr_value("href"))

    i = i + 1
end