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(Go) Iterate over Direct Child Nodes by Index

Demonstrates some ways to iterate over direct child nodes by index.

The input XML, available at http://www.chilkatsoft.com/data/get_child.xml, is this:

<root>
    <fruit color="red">apple</fruit>
    <fruit color="green">pear</fruit>
    <veg color="orange">carrot</veg>
    <meat animal="cow">beef</meat>
    <xyz>
        <fruit color="blue">blueberry</fruit>
        <veg color="green">broccoli</veg>
    </xyz>
    <fruit color="purple">grape</fruit>
    <cheese color="yellow">cheddar</cheese>
</root>

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    xml := chilkat.NewXml()
    // child is a Xml

    var success bool
    // The sample input XML is available at http://www.chilkatsoft.com/data/get_child.xml
    success = xml.LoadXmlFile("get_child.xml")
    if success != true {
        fmt.Println(xml.LastErrorText())
        xml.DisposeXml()
        return
    }

    // The NumChildren property contains the number of direct
    // child nodes.  Note: The child nodes under "xyz" are NOT
    // direct children of "root".  Therefore, the "root" node has
    // 7 direct children
    fmt.Println("NumChildren = ", xml.NumChildren())

    // Iterate over the direct children by index. The first child
    // is at index 0.
    var i int
    for i = 0; i <= xml.NumChildren() - 1; i++ {
        //  access the tag and content directly by index:
        fmt.Println(i, ": ", *xml.GetChildTagByIndex(i), " : ", *xml.GetChildContentByIndex(i))

    }

    fmt.Println("-----")

    // Do the same as the above loop, but get the child node
    // and access the Tag and Content properties:
    for i = 0; i <= xml.NumChildren() - 1; i++ {
        child := xml.GetChild(i)
        fmt.Println(i, ": ", child.Tag(), " : ", child.Content())
        child.DisposeXml()
    }

    fmt.Println("-----")

    // Do the same as the above loop, but instead of creating
    // a new object instance for each child, call GetChild2 to
    // update the object's reference instead.
    for i = 0; i <= xml.NumChildren() - 1; i++ {
        // Navigate to the Nth child.  
        success = xml.GetChild2(i)
        fmt.Println(i, ": ", xml.Tag(), " : ", xml.Content())
        // Navigate back up to the parent:
        success = xml.GetParent2()
    }

    fmt.Println("-----")

    // Notice that the Content of the "xyz" node is empty.  This is correct.
    // The "xyz" node has no Content, but 2 direct children

    xml.DisposeXml()

 

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