(Go) Parsing XML Children Having Identical Tags

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

Chilkat Go Downloads Go Package for Windows, MacOS, Linux, Alpine Linux, Solaris

    // 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()