How Can I Directly Unmarshal XML into a Go Map for Optimized Data Processing?

Directly Unmarshal XML into a Map

Unmarshalling XML directly into a map can optimize data processing by eliminating the intermediate struct conversion step.

Implementation:

To achieve this, create a type that implements the xml.Unmarshaller interface, allowing direct mapping of XML data into a map[string]string.

type classAccessesMap struct {
    m map[string]string
}

// UnmarshalXML implements the xml.Unmarshaller interface
func (c *classAccessesMap) UnmarshalXML(d *xml.Decoder, start xml.StartElement) error {
    c.m = make(map[string]string)

    var key, val string

    for {
        t, _ := d.Token()
        switch tt := t.(type) {

            // TODO: parse the inner structure

        case xml.StartElement:
            fmt.Println(">", tt)

        case xml.EndElement:
            fmt.Println("<", tt)
            if tt.Name == start.Name {
                return nil
            }

            if tt.Name.Local == "enabled" {
                c.m[key] = val
            }
        }
    }
}

Usage:

Apply the custom type to the unmarshal operation to directly map the XML data into a map:

// Unmarshalling into a map
var cm classAccessesMap
err := xml.Unmarshal([]byte(`<classAccesses>\n    <apexClass>AccountRelationUtility</apexClass>\n    <enabled>true</enabled>\n</classAccesses>`), &cm)
if err != nil {
    fmt.Println(err)
}

This approach streamlines data processing by avoiding the overhead of dual conversions, resulting in improved performance for large data sets.

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