dphs48626 2016-07-13 17:50
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进行XML解析:使用属性作为结构字段名称

How can I use XML attributes as struct field?

This is my test: Each row correspond to a Person

package main

import (
    "encoding/xml"
    "fmt"
)

var xmlstr = `<data>
    <row>
        <col name='firstname'>John</col>
        <col name='age'>2</col>
    </row>
    <row>
        <col name='firstname'>3</col>
        <col name='age'>4</col>
    </row>
</data>`

type Data struct {
    XMLName xml.Name `xml:"data"`
    Person  []Person `xml:"row"`
}

type Person struct {
    PersonField []PersonField `xml:"col"`
}

type PersonField struct {
    Name  string `xml:"name,attr"`
    Value string `xml:",chardata"`
}

func main() {
    b := []byte(xmlstr)

    var d Data
    xml.Unmarshal(b, &d)

    for _, person := range d.Person {
        fmt.Println(person)
    }
}

I go a slice of 2 struct:

{[{firstname John} {age 2}]}
{[{firstname 3} {age 4}]}

How can I get this struct instead ? Where xml attributes is use as a struct field name?

type Person struct {
    Firstname string
    Age       int
}
  • 写回答

1条回答 默认 最新

  • dongxiang7276 2016-07-13 21:25
    关注

    You can define a custom unmarshaler for the <row> elements that does this "unpacking" for you:

    package main
    
    import (
        "encoding/xml"
        "fmt"
        "strconv"
    )
    
    var xmlstr = `<data>
        <row>
            <col name='firstname'>John</col>
            <col name='age'>2</col>
        </row>
        <row>
            <col name='firstname'>3</col>
            <col name='age'>4</col>
        </row>
    </data>`
    
    type Data struct {
        XMLName xml.Name `xml:"data"`
        Person  []Person `xml:"row"`
    }
    
    type Person struct {
        Firstname string
        Age       int
    }
    
    func (p *Person) UnmarshalXML(d *xml.Decoder, start xml.StartElement) error {
        x := struct {
            Col []struct {
                Name  string `xml:"name,attr"`
                Value string `xml:",chardata"`
            } `xml:"col"`
        }{}
        err := d.DecodeElement(&x, &start)
        if err != nil {
            return err
        }
        for _, col := range x.Col {
            switch col.Name {
            case "firstname":
                p.Firstname = col.Value
            case "age":
                p.Age, err = strconv.Atoi(col.Value)
                if err != nil {
                    return err
                }
            }
        }
        return nil
    }
    
    func main() {
        b := []byte(xmlstr)
    
        var d Data
        if err := xml.Unmarshal(b, &d); err != nil {
            panic(err)
        }
    
        for _, person := range d.Person {
            fmt.Println(person)
        }
    }
    

    https://play.golang.org/p/DRF5axeBc0

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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