Go

XML Path Performance Optimizations

See more XML Examples

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

    success := false

    xml := chilkat.NewXml()

    // 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>
    // 
    success = xml.LoadXmlFile("qa_output/large.xml")
    if success != true {
        fmt.Println(xml.LastErrorText())
        xml.DisposeXml()
        return
    }

    // Iterating over the individual "license" nodes with this code snippet is
    // extremely slow:
    licCount := xml.NumChildrenHavingTag("licenses|license")
    fmt.Println("license count = ", licCount)

    var s *string = new(string)
    i := 0
    // If "10" is changed to licCount, then it becomes apparent that this loop gets slower with each iteration.
    for i < 10 {
        xml.SetI(i)
        s = xml.GetChildContent("licenses|license[i]|id")
        fmt.Println(i, ": ", *s)
        i = i + 1
    }

    // 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.
    for i < 10 {
        xml.SetI(i)
        s = xml.GetChildContent("licenses|[i]|id")
        fmt.Println(i, ": ", *s)
        i = i + 1
    }

    // 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.NumChildrenAt("licenses")
    fmt.Println("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

    xml.DisposeXml()