Can XML be Unmarhaled Directly into a Go Map to Improve Performance?

Unmarshaling XML Directly Into a Map

Problem:

Unmarshalling XML into a struct and then converting it into a map is time-consuming for large datasets. Is there a way to unmarshal directly into a map?

XML:

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

Current Struct:

type classAccesses struct {
    ApexClass string `xml:"apexClass"`
    Enabled   string `xml:"enabled"`
}

type diffs struct {
    ClassAccesses []classAccesses `xml:"classAccesses"`
}

Desired Map:

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

Solution:

Implement the xml.Unmarshaller interface to marshal data directly into a map[string]string.

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) {

        // 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
            }
        }
    }
}

Partial Solution:

Visit https://play.golang.org/p/7aOQ5mcH6zQ for a partial implementation.

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