How Can I Efficiently Unmarshal XML Directly into a Go Map[string]string?

Unmarshal XML Directly Into a Map

When dealing with large XML datasets, the dual conversion process of unmarshaling into a struct and then into a map can become time-consuming. This article explores a more efficient approach to directly unmarshal XML into a map.

Problem:

The goal is to convert the following XML into a map[string]string, where the key-value pairs are extracted from child elements:

<classAccesses>
    <apexClass>AccountRelationUtility</apexClass>
    <enabled>true</enabled>
</classAccesses>

To achieve this, a custom data structure that implements the xml.Unmarshaller interface is required.

Solution:

type classAccessesMap struct {
    m map[string]string
}

func (c *classAccessesMap) UnmarshalXML(d *xml.Decoder, start xml.StartElement) error {
    c.m = map[string]string{}

    key := ""
    val := ""

    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
            }
        }
    }
}

By implementing the xml.Unmarshaller interface, the custom data structure can directly unmarshal the XML into a map[string]string. This eliminates the need for the intermediate struct conversion, resulting in a more efficient unmarshaling process.

Partial Solution and Demonstration:

A partial solution demonstrating the approach is available at https://play.golang.org/p/7aOQ5mcH6zQ. This solution includes the custom data structure and a sample XML to demonstrate the unmarshaling process.

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