How Can I Efficiently Unmarshal XML Directly into a Go Map?-Golang-php.cn

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

Barbara Streisand

Release： 2024-12-18 01:53:10

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

Unmarshaling XML Directly into a Map

Unmarshaling XML data into an intermediate struct that is then converted into a map can be time-consuming for large data sets. In such cases, direct unmarshaling into a map is a more efficient approach.

To unmarshal XML directly into a map, you can create a custom type that implements the xml.Unmarshaler interface. This type will handle the unmarshaling process and store the data in a map[string]string.

Example:

type classAccessesMap struct {
    m map[string]string
}

// UnmarshalXML implements the xml.Unmarshaler interface to unmarshal XML directly into the map.
func (c *classAccessesMap) UnmarshalXML(d *xml.Decoder, start xml.StartElement) error {
    c.m = map[string]string{}

    key := ""
    val := ""

    // Iteratively parse XML tokens.
    for {
        t, _ := d.Token()
        switch tt := t.(type) {

        // TODO: Handle the inner structure parsing here.

        case xml.StartElement:
            key = tt.Name.Local

        case xml.EndElement:
            // Store the key-value pair in the map when the end of the "enabled" element is reached.
            if tt.Name.Local == "enabled" {
                c.m[key] = val
            }

            // Return nil when the end of the "classAccesses" element is reached.
            if tt.Name == start.Name {
                return nil
            }
        }
    }
}

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Usage:

// Unmarshal the XML into the custom classAccessesMap type.
var classAccessesMap classAccessesMap
if err := xml.Unmarshal([]byte(xmlData), &classAccessesMap); err != nil {
    // Handle error
}

fmt.Println(classAccessesMap.m) // Prints the map containing the parsed data.

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