dongye1143 2014-08-27 15:09
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Golang-解组额外的XML属性

Is there a way to unmarshal XML tags with dynamic attributes (I don't know which attributes I'll get every time).

Maybe it's not supported yet. See Issue 3633: encoding/xml: support for collecting all attributes

Something like :

package main

import (
    "encoding/xml"
    "fmt"
)

func main() {
    var v struct {
        Attributes []xml.Attr `xml:",any"`
    }

    data := `<TAG ATTR1="VALUE1" ATTR2="VALUE2" />`

    err := xml.Unmarshal([]byte(data), &v)
    if err != nil {
        panic(err)
    }
    fmt.Println(v)
}
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2条回答 默认 最新

  • dpo15099 2014-08-27 18:08
    关注

    You need to implement your own XMLUnmarshaler

    package main
    
    import (
        "encoding/xml"
        "fmt"
    )
    
    type CustomTag struct {
        Name       string
        Attributes []xml.Attr
    }
    
    func (c *CustomTag) UnmarshalXML(d *xml.Decoder, start xml.StartElement) error {
        c.Name = start.Name.Local
        c.Attributes = start.Attr
        return d.Skip()
    }
    
    func main() {
        v := &CustomTag{}
    
        data := []byte(`<tag ATTR1="VALUE1" ATTR2="VALUE2" />`)
    
        err := xml.Unmarshal(data, &v)
        if err != nil {
            panic(err)
        }
        fmt.Printf("%+v
    ", v)
    }
    

    outputs

    &{Name:tag Attributes:[{Name:{Space: Local:ATTR1} Value:VALUE1} {Name:{Space: Local:ATTR2} Value:VALUE2}]}
    

    http://play.golang.org/p/9ZrpIT32o_

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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