How to Directly Unmarshal XML into a Go Map?-Golang-php.cn

How to Directly Unmarshal XML into a Go Map?

Linda Hamilton

Release： 2024-12-10 18:08:11

Original

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How to Directly Unmarshal XML into a Go Map?

Unmarshal XML Directly into a Map

Problem:

You need to unmarshal XML data into a map instead of first unmarshalling into a struct and then converting the struct into a map, which is taking too much time for a large dataset.

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

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Desired Map:

map[string]string{
    "ApexClass": "enabled"
}

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Solution:

Implement the xml.Unmarshaller interface for a custom type that represents the desired map structure:

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: ...

        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
            }
        }
    }
}

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Then, use the xml.Unmarshal function to parse the XML data directly into the custom type:

var classAccesses classAccessesMap
if err := xml.Unmarshal(data, &classAccesses); err != nil {
    // Handle error
}

fmt.Println(classAccesses.m)

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