Question

如何将XML属性用作struct field？

这是我的测试：每行对应一个人

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)
    }
}

我选择了2个结构：

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

我怎样才能获得这个结构？其中xml属性用作结构字段名称？

type Person struct {
    Firstname string
    Age       int
}

Answer 1

您可以为执行此操作的<row>元素定义自定义解组，并且＃34;解包＆＃34;为你：

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

Go XML解析：使用属性作为结构字段名称

1 个答案: