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Parsing XML Children Having Identical Tags

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

    success := false

    // 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>
    // 

    xml := chilkat.NewXml()

    success = xml.LoadXmlFile("qa_data/xml/georgeCostanza.xml")

    numPhoneNumbers := xml.NumChildrenHavingTag("entry|*:phoneNumber")
    i := 0
    for i < numPhoneNumbers {
        xPhoneNumber := xml.GetNthChildWithTag("entry|*:phoneNumber",i)
        fmt.Println(xPhoneNumber.Content())
        xPhoneNumber.DisposeXml()
        i = i + 1
    }

    fmt.Println("----")

    numGroupMemberships := xml.NumChildrenHavingTag("entry|*:groupMembershipInfo")
    i = 0
    for i < numGroupMemberships {
        xMembership := xml.GetNthChildWithTag("entry|*:groupMembershipInfo",i)
        fmt.Println(*xMembership.GetAttrValue("href"))
        xMembership.DisposeXml()
        i = i + 1
    }


    xml.DisposeXml()